## 60 номер – Д 1853

### Пример:

$\int \dfrac{xdx}{\sqrt{ 5+x-x^{2} }}$

### Решение:

$f(x)=5+x-x^{2}$
$f'(x)=1-2x$

$-\dfrac12 f'(x)=x-\dfrac12$
$x=-\dfrac12 f'(x)+\dfrac12$

$I=\int \dfrac{x}{\sqrt{f(x)}}dx=-\dfrac12\int \dfrac{f'(x)}{\sqrt{f(x)}}dx+\dfrac12\int \dfrac{dx}{\sqrt{f(x)}}$

$I=I_1+I_2$

$u=f(x);\ du=f'(x)dx$
$I_1=-\dfrac12\int \dfrac{du}{\sqrt{u}}=-\sqrt{u}=-\sqrt{f(x)}$

$f(x)=5+x-x^2=\dfrac{21}{4}-(x-\dfrac12)^2$

$t=x-\dfrac12;\ dt=dx$
$I_2=\dfrac12\int \dfrac{dt}{\sqrt{\dfrac{21}{4}-t^2}}=\dfrac12\arcsin(\dfrac{t}{\sqrt{21}/2})=\dfrac12\arcsin(\dfrac{2x-1}{\sqrt{21}})$

${\,I=-\sqrt{5+x-x^{2}}+\dfrac12\arcsin(\dfrac{2x-1}{\sqrt{21}})+C\,}$

## 61 номер – Д 1903

### Пример:

$\int \dfrac{x^{3}}{(x-1)^{100}}dx$

### Решение:

$I=\int \dfrac{x^{3}}{(x-1)^{100}}dx$

$t=x-1;\ x=t+1;\ dt=dx$
$I=\int \dfrac{(t+1)^{3}}{t^{100}}dt=\int \dfrac{t^{3}+3t^{2}+3t+1}{t^{100}}dt=\int (t^{-97}+3t^{-98}+3t^{-99}+t^{-100})dt$

$I=-\dfrac{1}{96}t^{-96}-\dfrac{3}{97}t^{-97}-\dfrac{3}{98}t^{-98}-\dfrac{1}{99}t^{-99}+C$

${\,I=-\dfrac{1}{96(x-1)^{96}}-\dfrac{3}{97(x-1)^{97}}-\dfrac{3}{98(x-1)^{98}}-\dfrac{1}{99(x-1)^{99}}+C\,}$

## 62 номер – Д 1905

### Пример:

$\int \dfrac{x^{3}dx}{x^{8}+3}$

### Решение:

$I=\int \dfrac{x^{3}}{x^{8}+3}dx$

$t=x^{4};\ dt=4x^{3}dx;\ x^{3}dx=\dfrac14dt$
$I=\dfrac14\int \dfrac{dt}{t^{2}+3}=\dfrac14\cdot\dfrac{1}{\sqrt3}\arctan(\dfrac{t}{\sqrt3})+C$

${\,I=\dfrac{1}{4\sqrt3}\arctan(\dfrac{x^{4}}{\sqrt3})+C\,}$

## 63 номер – Д 1907

### Пример:

$\int \dfrac{x^{4}-3}{x(x^{8}+3x^{4}+2)}$

### Решение:

$I=\int \dfrac{x^{4}-3}{x(x^{8}+3x^{4}+2)}dx$
$x^{8}+3x^{4}+2=(x^{4})^{2}+3x^{4}+2=(x^{4}+1)(x^{4}+2)$

$t=x^{4};\ dt=4x^{3}dx;\ \dfrac{dt}{t}=4\dfrac{dx}{x};\ \dfrac{dx}{x}=\dfrac{1}{4}\dfrac{dt}{t}$
$I=\int \dfrac{x^{4}-3}{x(x^{4}+1)(x^{4}+2)}dx=\int \dfrac{t-3}{(t+1)(t+2)}\dfrac{dx}{x}=\dfrac14\int \dfrac{t-3}{t(t+1)(t+2)}dt$

$\dfrac{t-3}{t(t+1)(t+2)}=\dfrac{A}{t}+\dfrac{B}{t+1}+\dfrac{C}{t+2}$
$t-3=A(t+1)(t+2)+Bt(t+2)+Ct(t+1)$
$A=-\dfrac{3}{2};\ B=4;\ C=-\dfrac{5}{2}$

$I=\dfrac14\int(-\dfrac{3}{2}\dfrac{1}{t}+4\dfrac{1}{t+1}-\dfrac{5}{2}\dfrac{1}{t+2})dt$
$I=-\dfrac{3}{8}\ln|t|+\ln|t+1|-\dfrac{5}{8}\ln|t+2|+C$
$\ln|t|=\ln(x^{4})=4\ln|x|$

${\,I=\ln(x^{4}+1)-\dfrac{3}{2}\ln|x|-\dfrac{5}{8}\ln(x^{4}+2)+C\,}$

## 64 номер – Д 1909

### Пример:

$\int \dfrac{x^{11}dx}{x^{8}+3x^{4}+2}$

### Решение:

$I=\int \dfrac{x^{11}}{x^{8}+3x^{4}+2}dx$

$t=x^{4};\ dt=4x^{3}dx;\ x^{11}dx=x^{8}\cdot x^{3}dx=t^{2}\cdot\dfrac14dt$
$x^{8}+3x^{4}+2=t^{2}+3t+2=(t+1)(t+2)$

$I=\dfrac14\int \dfrac{t^{2}}{t^{2}+3t+2}dt=\dfrac14\int(1-\dfrac{3t+2}{(t+1)(t+2)})dt$

$\dfrac{3t+2}{(t+1)(t+2)}=\dfrac{A}{t+1}+\dfrac{B}{t+2}$
$3t+2=A(t+2)+B(t+1)=(A+B)t+(2A+B)$
$A=-1;\ B=4$

$I=\dfrac14\int(1+\dfrac{1}{t+1}-\dfrac{4}{t+2})dt=\dfrac14(t+\ln(t+1)-4\ln(t+2))+C$

${\,I=\dfrac{x^{4}}{4}+\dfrac14\ln(x^{4}+1)-\ln(x^{4}+2)+C\,}$

## 65 номер – Д 1910

### Пример:

$\int \dfrac{x^{9}dx}{(x^{10}+2x^{5}+2)^{2}}$

### Решение:

$I=\int \dfrac{x^{9}}{(x^{10}+2x^{5}+2)^{2}}dx$

$t=x^{5};\ dt=5x^{4}dx;\ x^{9}dx=x^{5}\cdot x^{4}dx=t\cdot\dfrac15dt$
$x^{10}+2x^{5}+2=t^{2}+2t+2$

$I=\dfrac15\int \dfrac{t}{(t^{2}+2t+2)^{2}}dt=\dfrac15\int(\dfrac{t+1}{(t^{2}+2t+2)^{2}}-\dfrac{1}{(t^{2}+2t+2)^{2}})dt$
$I=\dfrac15(I_1-I_2)$

$u=t^{2}+2t+2;\ du=(2t+2)dt=2(t+1)dt$
$I_1=\int \dfrac{t+1}{(t^{2}+2t+2)^{2}}dt=\dfrac12\int \dfrac{du}{u^{2}}=-\dfrac{1}{2u}=-\dfrac{1}{2(t^{2}+2t+2)}$

$t^{2}+2t+2=(t+1)^{2}+1$
$s=t+1;\ ds=dt$
$I_2=\int \dfrac{dt}{(t^{2}+2t+2)^{2}}=\int \dfrac{ds}{(s^{2}+1)^{2}}$

$(\dfrac{s}{s^{2}+1})'=\dfrac{1-s^{2}}{(s^{2}+1)^{2}}$
$\dfrac{1}{(s^{2}+1)^{2}}=\dfrac{1-s^{2}}{(s^{2}+1)^{2}}+\dfrac{s^{2}}{(s^{2}+1)^{2}}=(\dfrac{s}{s^{2}+1})'+(\dfrac{1}{s^{2}+1}-\dfrac{1}{(s^{2}+1)^{2}})$
$2\int \dfrac{ds}{(s^{2}+1)^{2}}=\dfrac{s}{s^{2}+1}+\arctan s$
$I_2=\dfrac12(\dfrac{s}{s^{2}+1}+\arctan s)=\dfrac12(\dfrac{t+1}{t^{2}+2t+2}+\arctan(t+1))$

$I=\dfrac15(-\dfrac{1}{2(t^{2}+2t+2)}-\dfrac12(\dfrac{t+1}{t^{2}+2t+2}+\arctan(t+1)))+C$
$I=-\dfrac{t+2}{10(t^{2}+2t+2)}-\dfrac{1}{10}\arctan(t+1)+C$

${\,I=-\dfrac{x^{5}+2}{10(x^{10}+2x^{5}+2)}-\dfrac{1}{10}\arctan(x^{5}+1)+C\,}$

## 66 номер – Д 1913

### Пример:

$\int \dfrac{dx}{x(x^{10}+2)}$

### Решение:

$I=\int \dfrac{dx}{x(x^{10}+2)}$

$t=x^{10};\ dt=10x^{9}dx;\ dx=\dfrac{dt}{10x^{9}}$
$I=\int \dfrac{\dfrac{dt}{10x^{9}}}{x(t+2)}=\dfrac{1}{10}\int \dfrac{dt}{x^{10}(t+2)}=\dfrac{1}{10}\int \dfrac{dt}{t(t+2)}$

$\dfrac{1}{t(t+2)}=\dfrac{A}{t}+\dfrac{B}{t+2}$
$1=A(t+2)+Bt=(A+B)t+2A$
$A=\dfrac12;\ B=-\dfrac12$

$I=\dfrac{1}{10}\int(\dfrac{1}{2t}-\dfrac{1}{2(t+2)})dt=\dfrac{1}{20}(\ln|t|-\ln|t+2|)+C$

${\,I=\dfrac{1}{20}\ln|\dfrac{x^{10}}{x^{10}+2}|+C\,}$

## 67 номер – Д 1915

### Пример:

$\int \dfrac{1-x^{7}}{x(1+x^{7})} \, dx$

### Решение:

$I=\int \dfrac{1-x^{7}}{x(1+x^{7})}dx$

$\dfrac{1-x^{7}}{x(1+x^{7})}=\dfrac{1+x^{7}}{x(1+x^{7})}-\dfrac{2x^{7}}{x(1+x^{7})}=\dfrac{1}{x}-\dfrac{2x^{6}}{1+x^{7}}$

$I=\int \dfrac{dx}{x}-2\int \dfrac{x^{6}}{1+x^{7}}dx$

$u=1+x^{7};\ du=7x^{6}dx$
$\int \dfrac{x^{6}}{1+x^{7}}dx=\dfrac{1}{7}\int \dfrac{du}{u}=\dfrac{1}{7}\ln|u|$

$I=\ln|x|-\dfrac{2}{7}\ln|1+x^{7}|+C$

## 68 номер – Д 1916

### Пример:

$\int \dfrac{x^{4}-1}{x(x^{3}-5)(x^{5}-5x+1)} \, dx$

### Решение:

$I=\int \dfrac{x^{4}-1}{x(x^{3}-5)(x^{5}-5x+1)}dx$

$\dfrac{x^{4}-1}{x(x^{3}-5)(x^{5}-5x+1)}=\dfrac{1}{5x}+\dfrac{187x^{2}+1575x+1250}{7190(x^{3}-5)}-\dfrac{325x^{4}+315x^{3}+250x^{2}+187x-1488}{1438(x^{5}-5x+1)}$

$I=I_1+I_2+I_3$

$I_1=\dfrac15\int \dfrac{dx}{x}=\dfrac15\ln|x|$

$I_2=\dfrac{1}{7190}\int \dfrac{187x^{2}+1575x+1250}{x^{3}-5}dx=\dfrac{1}{7190}(\dfrac{187}{3}\int \dfrac{3x^{2}}{x^{3}-5}dx+\int \dfrac{1575x+1250}{x^{3}-5}dx)$
$u=x^{3}-5;\ du=3x^{2}dx$
$\dfrac{187}{3}\int \dfrac{3x^{2}}{x^{3}-5}dx=\dfrac{187}{3}\ln|x^{3}-5|$

$a=\sqrt[3]{5};\ x^{3}-5=(x-a)(x^{2}+ax+a^{2})$
$\dfrac{1575x+1250}{x^{3}-5}=\dfrac{A}{x-a}+\dfrac{Bx+C}{x^{2}+ax+a^{2}}$
$A=\dfrac{1575a+1250}{3a^{2}};\ B=-A;\ C=\dfrac{1575a-2500}{3a}$

$\int \dfrac{A}{x-a}dx=A\ln|x-a|$
$\int \dfrac{Bx+C}{x^{2}+ax+a^{2}}dx=\dfrac{B}{2}\ln(x^{2}+ax+a^{2})+(C-\dfrac{Ba}{2})\int \dfrac{dx}{x^{2}+ax+a^{2}}$
$x^{2}+ax+a^{2}=(x+\dfrac{a}{2})^{2}+\dfrac{3a^{2}}{4}$
$\int \dfrac{dx}{x^{2}+ax+a^{2}}=\dfrac{2}{a\sqrt3}\arctan(\dfrac{2x+a}{a\sqrt3})$

$I_2=\dfrac{1}{7190}(\dfrac{187}{3}\ln|x^{3}-5|+A\ln|x-a|+\dfrac{B}{2}\ln(x^{2}+ax+a^{2})+(C-\dfrac{Ba}{2})\dfrac{2}{a\sqrt3}\arctan(\dfrac{2x+a}{a\sqrt3}))$

$I_3=-\dfrac{1}{1438}\int \dfrac{325x^{4}+315x^{3}+250x^{2}+187x-1488}{x^{5}-5x+1}dx$
$P(x)=x^{5}-5x+1;\ P'(x)=5x^{4}-5$
$r_1,\dots,r_5\text{ — корни }P(x)=0$
$\dfrac{325x^{4}+315x^{3}+250x^{2}+187x-1488}{P(x)}=\sum\limits_{k=1}^{5}\dfrac{325r_k^{4}+315r_k^{3}+250r_k^{2}+187r_k-1488}{P'(r_k)}\cdot\dfrac{1}{x-r_k}$
$I_3=-\dfrac{1}{1438}\sum\limits_{k=1}^{5}\dfrac{325r_k^{4}+315r_k^{3}+250r_k^{2}+187r_k-1488}{5r_k^{4}-5}\ln(x-r_k)$

${\,I=\dfrac15\ln|x|+I_2+I_3+C\,}$

## 69 номер – Д 1917

### Пример:

$\int \dfrac{x^{2}+1}{x^{4}+x^{2}+1} \, dx$

### Решение:

$I=\int \dfrac{x^{2}+1}{x^{4}+x^{2}+1}dx$

$x^{4}+x^{2}+1=(x^{2}-x+1)(x^{2}+x+1)$
$x^{2}+1=\dfrac12[(x^{2}-x+1)+(x^{2}+x+1)]$

$I=\dfrac12\int \dfrac{dx}{x^{2}-x+1}+\dfrac12\int \dfrac{dx}{x^{2}+x+1}=I_1+I_2$

$x^{2}-x+1=(x-\dfrac12)^{2}+\dfrac34$
$t=x-\dfrac12;\ dt=dx$
$I_1=\dfrac12\int \dfrac{dt}{t^{2}+(\frac{\sqrt3}{2})^{2}}=\dfrac{1}{\sqrt3}\arctan(\dfrac{2t}{\sqrt3})=\dfrac{1}{\sqrt3}\arctan(\dfrac{2x-1}{\sqrt3})$

$x^{2}+x+1=(x+\dfrac12)^{2}+\dfrac34$
$s=x+\dfrac12;\ ds=dx$
$I_2=\dfrac12\int \dfrac{ds}{s^{2}+(\frac{\sqrt3}{2})^{2}}=\dfrac{1}{\sqrt3}\arctan(\dfrac{2s}{\sqrt3})=\dfrac{1}{\sqrt3}\arctan(\dfrac{2x+1}{\sqrt3})$

${\,I=\dfrac{1}{\sqrt3}(\arctan(\dfrac{2x-1}{\sqrt3})+\arctan(\dfrac{2x+1}{\sqrt3}))+C\,}$

## 70 номер – Д 1921

да может ну это... ну не надо?

## 71 номер – Д 1971

### Пример:

$\int \dfrac{dx}{\sqrt{ x^{2}+1 }-\sqrt{ x^{2}-1 }}$

### Решение:

$x^2-1\ge0;\ |x|\ge1;$

$I=\int \dfrac{dx}{\sqrt{x^2+1}-\sqrt{x^2-1}}\cdot\dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{\sqrt{x^2+1}+\sqrt{x^2-1}}=\int \dfrac{\sqrt{x^2+1}+\sqrt{x^2-1}}{(x^2+1)-(x^2-1)}dx$
$I=\dfrac12\int(\sqrt{x^2+1}+\sqrt{x^2-1})dx=\dfrac12\int\sqrt{x^2+1}\,dx+\dfrac12\int\sqrt{x^2-1}\,dx$
$I=I_1+I_2$

$I_1=\dfrac12\int\sqrt{x^2+1}\,dx;\ \sqrt{x^2+1}=r$
$\int r\,dx=\dfrac12(xr+\ln(x+r))$
$I_1=\dfrac12\cdot\dfrac12(x\sqrt{x^2+1}+\ln|x+\sqrt{x^2+1}|)=\dfrac14x\sqrt{x^2+1}+\dfrac14\ln|x+\sqrt{x^2+1}|$

$I_2=\dfrac12\int\sqrt{x^2-1}\,dx;\ \sqrt{x^2-1}=s$
$\int s\,dx=\dfrac12(xs-\ln|x+s|)$
$I_2=\dfrac12\cdot\dfrac12(x\sqrt{x^2-1}-\ln|x+\sqrt{x^2-1}|)=\dfrac14x\sqrt{x^2-1}-\dfrac14\ln|x+\sqrt{x^2-1}|$

${\,I=\dfrac{x}{4}(\sqrt{x^{2}+1}+\sqrt{x^{2}-1})+\dfrac14\ln|\dfrac{x+\sqrt{x^{2}+1}}{x+\sqrt{x^{2}-1}}|+C\,}$

## 72 номер – Д 1972

### Пример:

$\int \dfrac{xdx}{(1-x^{3})\sqrt{ 1-x^{2} }}$

### Решение:

$I=\int \dfrac{x}{(1-x^3)\sqrt{1-x^2}}dx$

$\dfrac{1}{1-x^3}=\dfrac{1}{3(1-x)}+\dfrac{x+2}{3(x^2+x+1)}$
$I=\dfrac13\int \dfrac{x}{(1-x)\sqrt{1-x^2}}dx+\dfrac13\int \dfrac{x(x+2)}{(x^2+x+1)\sqrt{1-x^2}}dx$
$I=I_1+I_2$

$x=\cos t;\ dx=-\sin t\,dt;\ \sqrt{1-x^2}=\sin t$
$I_1=\dfrac13\int \dfrac{\cos t}{(1-\cos t)\sin t}(-\sin t\,dt)=-\dfrac13\int \dfrac{\cos t}{1-\cos t}dt$
$\dfrac{\cos t}{1-\cos t}=-1+\dfrac{1}{1-\cos t}$
$I_1=-\dfrac13\int(-1+\dfrac{1}{1-\cos t})dt=\dfrac{t}{3}-\dfrac13\int\dfrac{dt}{1-\cos t}$
$1-\cos t=2\sin^2\dfrac{t}{2}$
$\int\dfrac{dt}{1-\cos t}=\int\dfrac{dt}{2\sin^2(t/2)}=-\cot\dfrac{t}{2}$
$I_1=\dfrac{t}{3}+\dfrac13\cot\dfrac{t}{2}$

$I_2=\dfrac13\int \dfrac{\cos t(\cos t+2)}{(\cos^2t+\cos t+1)\sin t}(-\sin t\,dt)=-\dfrac13\int \dfrac{\cos t(\cos t+2)}{\cos^2t+\cos t+1}dt$
$\cos t(\cos t+2)=\cos^2t+2\cos t=(\cos^2t+\cos t+1)+(\cos t-1)$
$I_2=-\dfrac13\int(1+\dfrac{\cos t-1}{\cos^2t+\cos t+1})dt=-\dfrac{t}{3}-\dfrac13\int \dfrac{\cos t-1}{\cos^2t+\cos t+1}dt$

$I=I_1+I_2=\dfrac13\cot\dfrac{t}{2}-\dfrac13\int \dfrac{\cos t-1}{\cos^2t+\cos t+1}dt$

$u=\tan\dfrac{t}{2};\ dt=\dfrac{2\,du}{1+u^2};\ \cos t=\dfrac{1-u^2}{1+u^2}$
$\cos t-1=-\dfrac{2u^2}{1+u^2}$
$\cos^2t+\cos t+1=\dfrac{u^4+3}{(1+u^2)^2}$
$\dfrac{\cos t-1}{\cos^2t+\cos t+1}dt=-\dfrac{4u^2}{u^4+3}du$

$I=\dfrac13\cot\dfrac{t}{2}+\dfrac{4}{3}\int\dfrac{u^2}{u^4+3}du$

$p=\sqrt[4]{3}$
$u^4+3=u^4+p^4=(u^2-\sqrt2\,pu+\sqrt3)(u^2+\sqrt2\,pu+\sqrt3)$
$\dfrac{u^2}{u^4+3}=\dfrac{\sqrt2\,p^3}{12}(\dfrac{u}{u^2-\sqrt2\,pu+\sqrt3}-\dfrac{u}{u^2+\sqrt2\,pu+\sqrt3})$

$\int\dfrac{u}{u^2+\sqrt2\,pu+\sqrt3}du=\dfrac12\ln(u^2+\sqrt2\,pu+\sqrt3)-\arctan(\dfrac{\sqrt2}{p}u+1)$
$\int\dfrac{u}{u^2-\sqrt2\,pu+\sqrt3}du=\dfrac12\ln(u^2-\sqrt2\,pu+\sqrt3)+\arctan(\dfrac{\sqrt2}{p}u-1)$

$\int\dfrac{u^2}{u^4+3}du=\dfrac{\sqrt2\,p^3}{24}\ln(\dfrac{u^2-\sqrt2\,pu+\sqrt3}{u^2+\sqrt2\,pu+\sqrt3})+\dfrac{\sqrt2\,p^3}{12}(\arctan(\dfrac{\sqrt2}{p}u-1)+\arctan(\dfrac{\sqrt2}{p}u+1))+C$

$\sqrt2\,p^3=\dfrac{3\sqrt2}{p}$
$I=\dfrac13\cot\dfrac{t}{2}+\dfrac{\sqrt2}{6p}\ln(\dfrac{u^2-\sqrt2\,pu+\sqrt3}{u^2+\sqrt2\,pu+\sqrt3})+\dfrac{\sqrt2}{3p}(\arctan(\dfrac{\sqrt2}{p}u-1)+\arctan(\dfrac{\sqrt2}{p}u+1))+C$

$\cot\dfrac{t}{2}=\dfrac{1+\cos t}{\sin t}=\dfrac{1+x}{\sqrt{1-x^2}}$
$u=\tan\dfrac{t}{2}=\dfrac{\sin t}{1+\cos t}=\dfrac{\sqrt{1-x^2}}{1+x}$

${\,I=\dfrac{1+x}{3\sqrt{1-x^2}}+\dfrac{\sqrt2}{6\sqrt[4]{3}}\ln(\dfrac{u^2-\sqrt2\sqrt[4]{3}\,u+\sqrt3}{u^2+\sqrt2\sqrt[4]{3}\,u+\sqrt3})+\dfrac{\sqrt2}{3\sqrt[4]{3}}(\arctan(\dfrac{\sqrt2}{\sqrt[4]{3}}u-1)+\arctan(\dfrac{\sqrt2}{\sqrt[4]{3}}u+1))+C\,}$
$u=\dfrac{\sqrt{1-x^2}}{1+x}$

## 73 номер – Д 1973

### Пример:

$\int \dfrac{dx}{\sqrt{ 2 }+\sqrt{ 1-x }+\sqrt{ 1+x }}$

### Решение:

$-1\le x\le1$

$I=\int \dfrac{dx}{\sqrt2+\sqrt{1-x}+\sqrt{1+x}}$

$x=\cos2t;\ dx=-2\sin2t\,dt$
$\sqrt{1-x}=\sqrt{1-\cos2t}=\sqrt2\sin t$
$\sqrt{1+x}=\sqrt{1+\cos2t}=\sqrt2\cos t$

$I=\int \dfrac{-2\sin2t\,dt}{\sqrt2(1+\sin t+\cos t)}=-\sqrt2\int \dfrac{\sin2t}{1+\sin t+\cos t}dt$

$(\sin t+\cos t)^2=1+2\sin t\cos t=1+\sin2t$
$\sin2t=(\sin t+\cos t)^2-1$

$\dfrac{\sin2t}{1+\sin t+\cos t}=\dfrac{(\sin t+\cos t)^2-1}{1+\sin t+\cos t}=\sin t+\cos t-1$

$I=-\sqrt2\int(\sin t+\cos t-1)dt=-\sqrt2(-\cos t+\sin t-t)+C$
$I=\sqrt2(\cos t-\sin t+t)+C$

$x=\cos2t\ \implies\ t=\dfrac12\arccos x$
$\cos t=\sqrt{\dfrac{1+\cos2t}{2}}=\sqrt{\dfrac{1+x}{2}};\ \sin t=\sqrt{\dfrac{1-\cos2t}{2}}=\sqrt{\dfrac{1-x}{2}}$

${\,I=\sqrt{1+x}-\sqrt{1-x}+\dfrac{1}{\sqrt2}\arccos x+C\,}$

## 74 номер – Д 1974

### Пример:

$\int \dfrac{x+\sqrt{ 1+x+x^{2} }}{1+x+\sqrt{ 1+x+x^{2} }} \, dx$

### Решение:

$I=\int \dfrac{x+\sqrt{1+x+x^{2}}}{1+x+\sqrt{1+x+x^{2}}}dx$
$S=\sqrt{1+x+x^{2}}$

$\dfrac{x+S}{1+x+S}=1-\dfrac{1}{1+x+S}$
$I=\int dx-\int\dfrac{dx}{1+x+S}=x-J$

$J=\int\dfrac{dx}{1+x+S}\cdot\dfrac{1+x-S}{1+x-S}=\int\dfrac{1+x-S}{(1+x)^{2}-S^{2}}dx$
$(1+x)^{2}-S^{2}=(1+2x+x^{2})-(1+x+x^{2})=x$
$J=\int\dfrac{1+x-S}{x}dx=\int(\dfrac{1}{x}+1-\dfrac{S}{x})dx=\ln|x|+x-K$

$I=x-(\ln|x|+x-K)=K-\ln|x|$
$K=\int\dfrac{S}{x}dx$

$S=xt+1;\ x\neq0$
$x^{2}+x+1=(xt+1)^{2}=x^{2}t^{2}+2xt+1$
$x+1=xt^{2}+2t$
$x(1-t^{2})=2t-1$
$x=\dfrac{2t-1}{1-t^{2}};\ S=xt+1$

$\dfrac{S}{x}=\dfrac{xt+1}{x}=t+\dfrac{1}{x}=t+\dfrac{1-t^{2}}{2t-1}=\dfrac{t^{2}-t+1}{2t-1}$
$dx=(\dfrac{2t-1}{1-t^{2}})'dt=\dfrac{2(t^{2}-t+1)}{(t^{2}-1)^{2}}dt$

$K=\int\dfrac{S}{x}dx=\int\dfrac{t^{2}-t+1}{2t-1}\cdot\dfrac{2(t^{2}-t+1)}{(t^{2}-1)^{2}}dt=\int\dfrac{2(t^{2}-t+1)^{2}}{(2t-1)(t^{2}-1)^{2}}dt$

$\dfrac{2(t^{2}-t+1)^{2}}{(2t-1)(t^{2}-1)^{2}}=\dfrac{2}{2t-1}+\dfrac{1}{2(t+1)}-\dfrac{3}{2(t+1)^{2}}-\dfrac{1}{2(t-1)}+\dfrac{1}{2(t-1)^{2}}$

$K=\int(\dfrac{2}{2t-1}+\dfrac{1}{2(t+1)}-\dfrac{3}{2(t+1)^{2}}-\dfrac{1}{2(t-1)}+\dfrac{1}{2(t-1)^{2}})dt$
$K=\ln|2t-1|+\dfrac12\ln|t+1|-\dfrac12\ln|t-1|+\dfrac{3}{2(t+1)}-\dfrac{1}{2(t-1)}+C$

$t=\dfrac{S-1}{x}=\dfrac{\sqrt{1+x+x^{2}}-1}{x}$
$2t-1=\dfrac{2S-x-2}{x}$
$t+1=\dfrac{S+x-1}{x}$
$t-1=\dfrac{S-x-1}{x}$

$I=K-\ln|x|$
$I=\ln|\dfrac{2S-x-2}{x}|+\dfrac12\ln|\dfrac{S+x-1}{S-x-1}|+\dfrac{3}{2}\cdot\dfrac{x}{S+x-1}-\dfrac{1}{2}\cdot\dfrac{x}{S-x-1}-\ln|x|+C$

${\,I=\ln|\dfrac{2\sqrt{1+x+x^{2}}-x-2}{x^{2}}|+\dfrac12\ln|\dfrac{\sqrt{1+x+x^{2}}+x-1}{\sqrt{1+x+x^{2}}-x-1}|+\dfrac{3x}{2(\sqrt{1+x+x^{2}}+x-1)}-\dfrac{x}{2(\sqrt{1+x+x^{2}}-x-1)}+C\,}$

## 75 номер – Д 1975

### Пример:

$\int \dfrac{\sqrt{ x(x+1) }}{\sqrt{ x }+\sqrt{ x+1 }} \, dx$

### Решение:

$I=\int \dfrac{\sqrt{x(x+1)}}{\sqrt{x}+\sqrt{x+1}}dx=\int \dfrac{\sqrt{x}\sqrt{x+1}}{\sqrt{x}+\sqrt{x+1}}dx$

$I=\int \dfrac{\sqrt{x}\sqrt{x+1}}{\sqrt{x}+\sqrt{x+1}}\cdot\dfrac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x+1}-\sqrt{x}}dx=\int \dfrac{\sqrt{x}\sqrt{x+1}(\sqrt{x+1}-\sqrt{x})}{(\sqrt{x+1})^{2}-(\sqrt{x})^{2}}dx$
$I=\int \sqrt{x}\sqrt{x+1}(\sqrt{x+1}-\sqrt{x})dx$

$I=\int(\sqrt{x}(x+1)-x\sqrt{x+1})dx=\int (x+1)\sqrt{x}\,dx-\int x\sqrt{x+1}\,dx$
$I=I_1-I_2$

$I_1=\int (x+1)\sqrt{x}\,dx=\int (x^{\frac32}+x^{\frac12})dx=\dfrac{2}{5}x^{\frac52}+\dfrac{2}{3}x^{\frac32}$

$I_2=\int x\sqrt{x+1}\,dx$
$t=x+1;\ x=t-1;\ dt=dx$
$I_2=\int (t-1)\sqrt{t}\,dt=\int (t^{\frac32}-t^{\frac12})dt=\dfrac{2}{5}t^{\frac52}-\dfrac{2}{3}t^{\frac32}=\dfrac{2}{5}(x+1)^{\frac52}-\dfrac{2}{3}(x+1)^{\frac32}$

${\,I=\dfrac{2}{5}x^{\frac52}+\dfrac{2}{3}x^{\frac32}-\dfrac{2}{5}(x+1)^{\frac52}+\dfrac{2}{3}(x+1)^{\frac32}+C\,}$

## 76 номер – Д 2025

### Пример:

$\int \dfrac{dx}{2\sin x-\cos x+5}$

### Решение:

$I=\int \dfrac{dx}{2\sin x-\cos x+5}$

$t=\tan\dfrac{x}{2};\ \sin x=\dfrac{2t}{1+t^2};\ \cos x=\dfrac{1-t^2}{1+t^2};\ dx=\dfrac{2dt}{1+t^2}$

$I=\int \dfrac{\dfrac{2dt}{1+t^2}}{2\cdot\dfrac{2t}{1+t^2}-\dfrac{1-t^2}{1+t^2}+5}=\int \dfrac{2dt}{4t-(1-t^2)+5(1+t^2)}=\int \dfrac{2dt}{6t^2+4t+4}=\int \dfrac{dt}{3t^2+2t+2}$

$3t^2+2t+2=3(t+\dfrac13)^2+\dfrac53$
$I=\int \dfrac{dt}{3(t+\dfrac13)^2+\dfrac53}=3\int \dfrac{dt}{9(t+\dfrac13)^2+5}$

$u=3t+1;\ du=3dt$
$I=\int \dfrac{du}{u^2+5}=\dfrac{1}{\sqrt5}\arctan(\dfrac{u}{\sqrt5})+C=\dfrac{1}{\sqrt5}\arctan(\dfrac{3\tan\frac{x}{2}+1}{\sqrt5})+C$

${\,I=\dfrac{1}{\sqrt5}\arctan(\dfrac{3\tan\frac{x}{2}+1}{\sqrt5})+C\,}$

## 77 номер – Д 2026

### Пример:

$\int \dfrac{dx}{(2+\cos x)\sin x}$

### Решение:

$I=\int \dfrac{dx}{(2+\cos x)\sin x}$

$t=\cos x;\ dt=-(\sin x)dx;\ dx=-\dfrac{dt}{\sin x};\ \sin^{2}x=1-\cos^{2}x=1-t^{2}$

$I=\int \dfrac{-\dfrac{dt}{\sin x}}{(2+t)\sin x}=-\int \dfrac{dt}{(t+2)\sin^{2}x}=-\int \dfrac{dt}{(t+2)(1-t^{2})}$
$1-t^{2}=(1-t)(1+t)$
$I=-\int \dfrac{dt}{(t+2)(1-t)(1+t)}$

$\dfrac{1}{(t+2)(1-t)(1+t)}=\dfrac{A}{t+2}+\dfrac{B}{1-t}+\dfrac{C}{1+t}$
$1=A(1-t)(1+t)+B(t+2)(1+t)+C(t+2)(1-t)$
$A=-\dfrac13;\ B=\dfrac16;\ C=\dfrac12$

$I=-\int(-\dfrac{1}{3(t+2)}+\dfrac{1}{6(1-t)}+\dfrac{1}{2(1+t)})dt$
$I=\int(\dfrac{1}{3(t+2)}-\dfrac{1}{6(1-t)}-\dfrac{1}{2(1+t)})dt$

$I=\dfrac13\ln|t+2|+\dfrac16\ln|1-t|-\dfrac12\ln|1+t|+C$

$t=\cos x$
${\,I=\dfrac13\ln|2+\cos x|+\dfrac16\ln|1-\cos x|-\dfrac12\ln|1+\cos x|+C\,}$

## 78 номер – Д 2027

### Пример:

$\int \dfrac{\sin ^{2}x}{\sin x+2\cos x} \, dx$

### Решение:

$I=\int \dfrac{\sin^2x}{\sin x+2\cos x}dx$

$\sin^2x=(\sin x+2\cos x)(\sin x-2\cos x)+4\cos^2x$
$\dfrac{\sin^2x}{\sin x+2\cos x}=\sin x-2\cos x+4\cdot\dfrac{\cos^2x}{\sin x+2\cos x}$

$I=\int(\sin x-2\cos x)dx+4\int\dfrac{\cos^2x}{\sin x+2\cos x}dx$
$I=-\cos x-2\sin x+4J$

$J=\int\dfrac{\cos^2x}{\sin x+2\cos x}dx$
$t=\tan\dfrac{x}{2};\ \sin x=\dfrac{2t}{1+t^2};\ \cos x=\dfrac{1-t^2}{1+t^2};\ dx=\dfrac{2dt}{1+t^2}$

$J=\int\dfrac{(\dfrac{1-t^2}{1+t^2})^2}{\dfrac{2t}{1+t^2}+2\cdot\dfrac{1-t^2}{1+t^2}}\cdot\dfrac{2dt}{1+t^2}=\int\dfrac{(1-t^2)^2}{(1+t^2)^2(1+t-t^2)}dt$

$\dfrac{(1-t^2)^2}{(1+t^2)^2(1+t-t^2)}=-\dfrac{4(t-2)}{5(t^2+1)^2}-\dfrac{4}{5(t^2+1)}-\dfrac{1}{5(t^2-t-1)}$

$J=-\dfrac45\int\dfrac{t-2}{(t^2+1)^2}dt-\dfrac45\int\dfrac{dt}{t^2+1}-\dfrac15\int\dfrac{dt}{t^2-t-1}$

$\int\dfrac{t-2}{(t^2+1)^2}dt=\int\dfrac{t}{(t^2+1)^2}dt-2\int\dfrac{dt}{(t^2+1)^2}$
$u=t^2+1;\ du=2t\,dt$
$\int\dfrac{t}{(t^2+1)^2}dt=\dfrac12\int\dfrac{du}{u^2}=-\dfrac{1}{2(t^2+1)}$

$(\dfrac{t}{t^2+1})'=\dfrac{1-t^2}{(t^2+1)^2}$
$2\int\dfrac{dt}{(t^2+1)^2}=\dfrac{t}{t^2+1}+\int\dfrac{dt}{t^2+1}=\dfrac{t}{t^2+1}+\arctan t$
$\int\dfrac{dt}{(t^2+1)^2}=\dfrac12(\dfrac{t}{t^2+1}+\arctan t)$

$\int\dfrac{t-2}{(t^2+1)^2}dt=-\dfrac{2t+1}{2(t^2+1)}-\arctan t$

$t^2-t-1=(t-\dfrac12)^2-(\dfrac{\sqrt5}{2})^2$
$\int\dfrac{dt}{t^2-t-1}=\dfrac{1}{\sqrt5}\ln|\dfrac{t-\frac12-\frac{\sqrt5}{2}}{t-\frac12+\frac{\sqrt5}{2}}|$

$J=-\dfrac45(-\dfrac{2t+1}{2(t^2+1)}-\arctan t)-\dfrac45\arctan t-\dfrac{1}{5\sqrt5}\ln|\dfrac{t-\frac12-\frac{\sqrt5}{2}}{t-\frac12+\frac{\sqrt5}{2}}|$
$J=\dfrac{4t+2}{5(t^2+1)}+\dfrac{\sqrt5}{25}\ln|\dfrac{2t-1+\sqrt5}{2t-1-\sqrt5}|+C$

$t=\tan\dfrac{x}{2}$
$I=-\dfrac{\cos x+2\sin x}{5}+\dfrac{4\sqrt5}{25}\ln|\dfrac{2\tan\frac{x}{2}-1+\sqrt5}{2\tan\frac{x}{2}-1-\sqrt5}|+C$

## 79 номер – Д 2029

### Пример:

$\int \dfrac{\sin ^{2}x}{1+\sin ^{2}x} \, dx$

### Решение:

$I=\int \dfrac{\sin^2x}{1+\sin^2x}dx$

$\dfrac{\sin^2x}{1+\sin^2x}=1-\dfrac{1}{1+\sin^2x}$
$I=\int dx-\int \dfrac{dx}{1+\sin^2x}=x-J$

$J=\int \dfrac{dx}{1+\sin^2x}$
$t=\tan x;\ dt=(1+t^2)dx;\ dx=\dfrac{dt}{1+t^2}$
$\sin^2x=\dfrac{\tan^2x}{1+\tan^2x}=\dfrac{t^2}{1+t^2}$

$J=\int \dfrac{\dfrac{dt}{1+t^2}}{1+\dfrac{t^2}{1+t^2}}=\int \dfrac{\dfrac{dt}{1+t^2}}{\dfrac{1+2t^2}{1+t^2}}=\int \dfrac{dt}{1+2t^2}$

$J=\dfrac{1}{\sqrt2}\arctan(\sqrt2\,t)+C=\dfrac{1}{\sqrt2}\arctan(\sqrt2\tan x)+C$

${\,I=x-\dfrac{1}{\sqrt2}\arctan(\sqrt2\tan x)+C\,}$

## 80 номер – Д 2030

### Пример:

$\int \dfrac{dx}{a^{2}\sin ^{2}x+b^{2}\cos ^{2}x}$

### Решение:

$I=\int \dfrac{dx}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}$

$I=\int \dfrac{dx}{\cos^{2}x(a^{2}\tan^{2}x+b^{2})}=\int \dfrac{\sec^{2}x}{a^{2}\tan^{2}x+b^{2}}dx$

$t=\tan x;\ dt=\sec^{2}x\,dx$
$I=\int \dfrac{dt}{a^{2}t^{2}+b^{2}}=\dfrac{1}{b^{2}}\int \dfrac{dt}{1+(\dfrac{a}{b})^{2}t^{2}}$

$I=\dfrac{1}{b^{2}}\cdot\dfrac{b}{a}\arctan(\dfrac{a}{b}t)+C=\dfrac{1}{ab}\arctan(\dfrac{a}{b}\tan x)+C$

${\,I=\dfrac{1}{ab}\arctan(\dfrac{a\tan x}{b})+C\,}$

## 81 номер – Д 2031

### Пример:

$\int \dfrac{\cos ^{2}xdx}{( a^{2}\sin ^{2} x+b^{2}\cos ^{2} x)^{2}}$

### Решение:

$I=\int \dfrac{\cos ^{2}x}{( a^{2}\sin ^{2} x+b^{2}\cos ^{2} x)^{2}}dx;\ (a\neq0,\ b\neq0)$

$t=\tan x;\ dt=(\tan x)'dx=\sec^2x\,dx=(1+\tan^2x)dx=(1+t^2)dx;\ dx=\dfrac{dt}{1+t^2};$
$\sin^2x=\dfrac{t^2}{1+t^2};\ \cos^2x=\dfrac{1}{1+t^2};$

$I=\int \dfrac{\dfrac{1}{1+t^2}\cdot\dfrac{dt}{1+t^2}}{(a^2\dfrac{t^2}{1+t^2}+b^2\dfrac{1}{1+t^2})^2}=\int \dfrac{\dfrac{dt}{(1+t^2)^2}}{(\dfrac{a^2t^2+b^2}{1+t^2})^2}=\int \dfrac{dt}{(a^2t^2+b^2)^2}$

$u=\dfrac{a}{b}t;\ t=\dfrac{b}{a}u;\ dt=\dfrac{b}{a}du;$
$a^2t^2+b^2=b^2(u^2+1);$

$I=\int \dfrac{\dfrac{b}{a}du}{(b^2(u^2+1))^2}=\dfrac{1}{ab^3}\int \dfrac{du}{(u^2+1)^2}$

$(\dfrac{u}{1+u^2})'=\dfrac{1-u^2}{(1+u^2)^2}$
$\dfrac{1}{(1+u^2)^2}=\dfrac12(\dfrac{1-u^2}{(1+u^2)^2}+\dfrac{1}{1+u^2})$

$\int \dfrac{du}{(1+u^2)^2}=\dfrac12\int \dfrac{1-u^2}{(1+u^2)^2}du+\dfrac12\int \dfrac{du}{1+u^2}=\dfrac12\cdot\dfrac{u}{1+u^2}+\dfrac12\arctan u$

$I=\dfrac{1}{ab^3}(\dfrac12\cdot\dfrac{u}{1+u^2}+\dfrac12\arctan u)=\dfrac{1}{2ab^3}(\dfrac{u}{1+u^2}+\arctan u)+C$

$u=\dfrac{a}{b}\tan x;$
$\dfrac{u}{1+u^2}=\dfrac{\frac{a}{b}\tan x}{1+(\frac{a}{b}\tan x)^2}=\dfrac{ab\tan x}{b^2+a^2\tan^2x}=\dfrac{ab\sin x\cos x}{a^2\sin^2x+b^2\cos^2x}$

${\,I=\dfrac{\sin x\cos x}{2b^{2}(a^{2}\sin ^{2}x+b^{2}\cos ^{2}x)}+\dfrac{1}{2ab^{3}}\arctan(\dfrac{a\tan x}{b})+C\,}$

## 82 номер – Д 2032

### Пример:

$\int \dfrac{\sin x\cos x}{\sin x+\cos x} \, dx$

### Решение:

$\sin x+\cos x\neq0;$

$I=\int \dfrac{\sin x\cos x}{\sin x+\cos x}dx$

$t=x-\dfrac{\pi}{4};\ x=t+\dfrac{\pi}{4};\ dt=dx$
$\sin x=\sin(t+\dfrac{\pi}{4})=\dfrac{\sin t+\cos t}{\sqrt2};$
$\cos x=\cos(t+\dfrac{\pi}{4})=\dfrac{\cos t-\sin t}{\sqrt2};$

$\sin x+\cos x=\dfrac{\sin t+\cos t+\cos t-\sin t}{\sqrt2}=\sqrt2\cos t;$
$\sin x\cos x=\dfrac{(\sin t+\cos t)(\cos t-\sin t)}{2}=\dfrac{\cos^2t-\sin^2t}{2}=\dfrac12\cos2t;$

$I=\int \dfrac{\frac12\cos2t}{\sqrt2\cos t}dt=\dfrac{1}{2\sqrt2}\int\dfrac{\cos2t}{\cos t}dt$
$\cos2t=2\cos^2t-1$
$\dfrac{\cos2t}{\cos t}=2\cos t-\sec t$

$I=\dfrac{1}{2\sqrt2}\int(2\cos t-\sec t)dt=\dfrac{1}{\sqrt2}\int\cos t\,dt-\dfrac{1}{2\sqrt2}\int\sec t\,dt$
$I=\dfrac{1}{\sqrt2}\sin t-\dfrac{1}{2\sqrt2}\ln|\sec t+\tan t|+C$

$\sin t=\sin(x-\dfrac{\pi}{4})=\dfrac{\sin x-\cos x}{\sqrt2};$
$\sec t+\tan t=\dfrac{1+\sin t}{\cos t};\ \sin t=\dfrac{\sin x-\cos x}{\sqrt2};\ \cos t=\cos(x-\dfrac{\pi}{4})=\dfrac{\sin x+\cos x}{\sqrt2};$
$\sec t+\tan t=\dfrac{\sqrt2+\sin x-\cos x}{\sin x+\cos x}$

${\,I=\dfrac{\sin x-\cos x}{2}-\dfrac{1}{2\sqrt2}\ln|\dfrac{\sqrt2+\sin x-\cos x}{\sin x+\cos x}|+C\,}$

## 83 номер – Д 2033

### Пример:

$\int \dfrac{dx}{(a\sin x+b\cos x)^{2}}$

### Решение:

$a^{2}+b^{2}\neq0;\ a\sin x+b\cos x\neq0;$

$I=\int \dfrac{dx}{(a\sin x+b\cos x)^{2}}$

$u=a\sin x+b\cos x;\ u'=a\cos x-b\sin x$
$v=a\cos x-b\sin x;\ v'=-a\sin x-b\cos x=-u$

$(\dfrac{v}{u})'=\dfrac{v'u-vu'}{u^{2}}=\dfrac{(-u)u-v^{2}}{u^{2}}=-\dfrac{u^{2}+v^{2}}{u^{2}}$

$u^{2}+v^{2}=(a\sin x+b\cos x)^{2}+(a\cos x-b\sin x)^{2}=a^{2}+b^{2}$

$(\dfrac{v}{u})'=-\dfrac{a^{2}+b^{2}}{(a\sin x+b\cos x)^{2}}$
$\dfrac{1}{(a\sin x+b\cos x)^{2}}=-\dfrac{1}{a^{2}+b^{2}}(\dfrac{v}{u})'$

$I=-\dfrac{1}{a^{2}+b^{2}}\int(\dfrac{v}{u})'dx=-\dfrac{1}{a^{2}+b^{2}}\cdot\dfrac{v}{u}+C$

${\,I=-\dfrac{a\cos x-b\sin x}{(a^{2}+b^{2})(a\sin x+b\cos x)}+C\,}$

## 84 номер – Д 2072

### Пример:

$\int x^{7}e^{ -x^{2} } \, dx$

### Решение:

$I=\int x^{7}e^{-x^{2}}dx$

$t=x^{2};\ dt=2x\,dx;\ x^{7}dx=x^{6}\cdot x\,dx=(x^{2})^{3}\cdot x\,dx=t^{3}\cdot\dfrac12dt$
$I=\dfrac12\int t^{3}e^{-t}dt$

$J=\int t^{3}e^{-t}dt;\ J=\int u\,dv;\ u=t^{3};\ dv=e^{-t}dt$
$du=3t^{2}dt;\ v=-e^{-t}$
$J=uv-\int v\,du=-t^{3}e^{-t}+3\int t^{2}e^{-t}dt$

$J_1=\int t^{2}e^{-t}dt;\ J_1=\int u\,dv;\ u=t^{2};\ dv=e^{-t}dt$
$du=2t\,dt;\ v=-e^{-t}$
$J_1=-t^{2}e^{-t}+2\int t e^{-t}dt$

$J_2=\int t e^{-t}dt;\ J_2=\int u\,dv;\ u=t;\ dv=e^{-t}dt$
$du=dt;\ v=-e^{-t}$
$J_2=-t e^{-t}+\int e^{-t}dt=-t e^{-t}-e^{-t}$

$J_1=-t^{2}e^{-t}+2(-t e^{-t}-e^{-t})=-(t^{2}+2t+2)e^{-t}$
$J=-t^{3}e^{-t}+3J_1=-t^{3}e^{-t}-3(t^{2}+2t+2)e^{-t}=-(t^{3}+3t^{2}+6t+6)e^{-t}$

$I=-\dfrac12(t^{3}+3t^{2}+6t+6)e^{-t}+C$
$t=x^{2}$

${\,I=-\dfrac12e^{-x^{2}}(x^{6}+3x^{4}+6x^{2}+6)+C\,}$

## 85 номер – Д 2073

### Пример:

$\int x^{2}e^{ \sqrt{ x } }dx$

### Решение:

$x\ge0$
$I=\int x^{2}e^{\sqrt{x}}dx$

$t=\sqrt{x};\ x=t^{2};\ dx=2t\,dt$
$I=\int t^{4}e^{t}\cdot2t\,dt=2\int t^{5}e^{t}dt$
$I=2J$

$J=\int t^{5}e^{t}dt;\ J=\int u\,dv;\ u=t^{5};\ dv=e^{t}dt$
$du=5t^{4}dt;\ v=e^{t}$
$J=t^{5}e^{t}-5\int t^{4}e^{t}dt=t^{5}e^{t}-5J_{1}$

$J_{1}=\int t^{4}e^{t}dt=t^{4}e^{t}-4\int t^{3}e^{t}dt=t^{4}e^{t}-4J_{2}$
$J_{2}=\int t^{3}e^{t}dt=t^{3}e^{t}-3\int t^{2}e^{t}dt=t^{3}e^{t}-3J_{3}$
$J_{3}=\int t^{2}e^{t}dt=t^{2}e^{t}-2\int te^{t}dt=t^{2}e^{t}-2J_{4}$
$J_{4}=\int te^{t}dt=te^{t}-\int e^{t}dt=te^{t}-e^{t}$

$J_{3}=t^{2}e^{t}-2(te^{t}-e^{t})=e^{t}(t^{2}-2t+2)$
$J_{2}=t^{3}e^{t}-3e^{t}(t^{2}-2t+2)=e^{t}(t^{3}-3t^{2}+6t-6)$
$J_{1}=t^{4}e^{t}-4e^{t}(t^{3}-3t^{2}+6t-6)=e^{t}(t^{4}-4t^{3}+12t^{2}-24t+24)$
$J=t^{5}e^{t}-5e^{t}(t^{4}-4t^{3}+12t^{2}-24t+24)=e^{t}(t^{5}-5t^{4}+20t^{3}-60t^{2}+120t-120)$

$I=2e^{t}(t^{5}-5t^{4}+20t^{3}-60t^{2}+120t-120)+C$
$t=\sqrt{x}$

${\,I=2e^{\sqrt{x}}(x^{2}\sqrt{x}-5x^{2}+20x\sqrt{x}-60x+120\sqrt{x}-120)+C\,}$

## 86 номер – Д 2074

### Пример:

$\int e^{ ax }\cos ^{2}bx \, dx$

### Решение:

$I=\int e^{ax}\cos^{2}(bx)\,dx$

$\cos^{2}(bx)=\dfrac{1+\cos(2bx)}{2}$
$I=\dfrac12\int e^{ax}dx+\dfrac12\int e^{ax}\cos(2bx)\,dx$
$I=I_1+I_2$

$a\neq0;$
$I_1=\dfrac12\int e^{ax}dx=\dfrac12\cdot\dfrac{e^{ax}}{a}=\dfrac{e^{ax}}{2a}$

$I_2=\dfrac12\int e^{ax}\cos(2bx)\,dx$
$\int e^{ax}\cos(kx)\,dx=\dfrac{e^{ax}}{a^{2}+k^{2}}(a\cos(kx)+k\sin(kx))$
$k=2b$
$I_2=\dfrac12\cdot\dfrac{e^{ax}}{a^{2}+4b^{2}}(a\cos(2bx)+2b\sin(2bx))$

${\,I=\dfrac{e^{ax}}{2a}+\dfrac{e^{ax}}{2(a^{2}+4b^{2})}(a\cos(2bx)+2b\sin(2bx))+C\,}$

$a=0;$
$I=\int \cos^{2}(bx)\,dx=\int\dfrac{1+\cos(2bx)}{2}dx=\dfrac{x}{2}+\dfrac{\sin(2bx)}{4b}+C;\ (b\neq0)$

## 87 номер – Д 2075

### Пример:

$\int e^{ ax }\sin ^{3}bx \, dx$

### Решение:

$I=\int e^{ax}\sin^{3}(bx)\,dx$

$\sin 3u=3\sin u-4\sin^{3}u$
$\sin^{3}u=\dfrac{3\sin u-\sin 3u}{4}$
$u=bx$
$\sin^{3}(bx)=\dfrac{3\sin(bx)-\sin(3bx)}{4}$

$I=\dfrac14\int e^{ax}(3\sin(bx)-\sin(3bx))dx=\dfrac34\int e^{ax}\sin(bx)dx-\dfrac14\int e^{ax}\sin(3bx)dx$
$I=\dfrac34 I_1-\dfrac14 I_2$

$\int e^{ax}\sin(kx)\,dx=\dfrac{e^{ax}}{a^{2}+k^{2}}(a\sin(kx)-k\cos(kx))+C$

$I_1=\int e^{ax}\sin(bx)\,dx=\dfrac{e^{ax}}{a^{2}+b^{2}}(a\sin(bx)-b\cos(bx))$
$I_2=\int e^{ax}\sin(3bx)\,dx=\dfrac{e^{ax}}{a^{2}+9b^{2}}(a\sin(3bx)-3b\cos(3bx))$

${\,I=\dfrac{e^{ax}}{4}(\dfrac{3(a\sin bx-b\cos bx)}{a^{2}+b^{2}}-\dfrac{a\sin 3bx-3b\cos 3bx}{a^{2}+9b^{2}})+C\,}$

## 88 номер – Д 2076

### Пример:

$\int xe^{ x }\sin x \, dx$

### Решение:

$I=\int xe^{x}\sin x\,dx$

$I=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x;\ du=dx$
$dv=e^{x}\sin x\,dx$

$v=\int e^{x}\sin x\,dx=\dfrac{e^{x}}{2}(\sin x-\cos x)$

$I=x\cdot\dfrac{e^{x}}{2}(\sin x-\cos x)-\int\dfrac{e^{x}}{2}(\sin x-\cos x)\,dx$
$I=\dfrac{x e^{x}}{2}(\sin x-\cos x)-\dfrac12\int e^{x}\sin x\,dx+\dfrac12\int e^{x}\cos x\,dx$

$\int e^{x}\sin x\,dx=\dfrac{e^{x}}{2}(\sin x-\cos x)$
$\int e^{x}\cos x\,dx=\dfrac{e^{x}}{2}(\sin x+\cos x)$

$I=\dfrac{x e^{x}}{2}(\sin x-\cos x)-\dfrac12\cdot\dfrac{e^{x}}{2}(\sin x-\cos x)+\dfrac12\cdot\dfrac{e^{x}}{2}(\sin x+\cos x)+C$
$I=\dfrac{x e^{x}}{2}(\sin x-\cos x)+\dfrac{e^{x}}{2}\cos x+C$

${\,I=\dfrac{e^{x}}{2}(x\sin x+(1-x)\cos x)+C\,}$

## 89 номер – Д 2077

### Пример:

$\int x^{2}e^{ x }\cos x \, dx$

### Решение:

$I=\int x^{2}e^{x}\cos x\,dx$

$I=\int u\,dv;\ \int u\,dv=uv-\int v\,du$
$u=x^{2};\ du=2x\,dx$
$dv=e^{x}\cos x\,dx$

$v=\int e^{x}\cos x\,dx=\dfrac{e^{x}}{2}(\sin x+\cos x)$

$I=\dfrac{x^{2}e^{x}}{2}(\sin x+\cos x)-\int x e^{x}(\sin x+\cos x)\,dx$
$I=\dfrac{x^{2}e^{x}}{2}(\sin x+\cos x)-J$

$J=\int x e^{x}(\sin x+\cos x)\,dx=\int x e^{x}\sin x\,dx+\int x e^{x}\cos x\,dx$
$J=J_{1}+J_{2}$

$J_{1}=\int x e^{x}\sin x\,dx$
$J_{1}=\int u\,dv;\ u=x;\ du=dx;\ dv=e^{x}\sin x\,dx$
$\int e^{x}\sin x\,dx=\dfrac{e^{x}}{2}(\sin x-\cos x)$
$J_{1}=x\cdot\dfrac{e^{x}}{2}(\sin x-\cos x)-\int\dfrac{e^{x}}{2}(\sin x-\cos x)\,dx$
$\int e^{x}\sin x\,dx=\dfrac{e^{x}}{2}(\sin x-\cos x);\ \int e^{x}\cos x\,dx=\dfrac{e^{x}}{2}(\sin x+\cos x)$
$J_{1}=\dfrac{x e^{x}}{2}(\sin x-\cos x)-\dfrac12\cdot\dfrac{e^{x}}{2}(\sin x-\cos x)+\dfrac12\cdot\dfrac{e^{x}}{2}(\sin x+\cos x)$
$J_{1}=\dfrac{e^{x}}{2}(x\sin x+(1-x)\cos x)$

$J_{2}=\int x e^{x}\cos x\,dx$
$J_{2}=\int u\,dv;\ u=x;\ du=dx;\ dv=e^{x}\cos x\,dx$
$J_{2}=x\cdot\dfrac{e^{x}}{2}(\sin x+\cos x)-\int\dfrac{e^{x}}{2}(\sin x+\cos x)\,dx$
$\int e^{x}(\sin x+\cos x)\,dx=\int e^{x}\sin x\,dx+\int e^{x}\cos x\,dx=e^{x}\sin x$
$J_{2}=\dfrac{x e^{x}}{2}(\sin x+\cos x)-\dfrac{e^{x}}{2}\sin x$

$J=\dfrac{e^{x}}{2}(x\sin x+(1-x)\cos x)+\dfrac{e^{x}}{2}(x(\sin x+\cos x)-\sin x)=\dfrac{e^{x}}{2}((2x-1)\sin x+\cos x)$

$I=\dfrac{x^{2}e^{x}}{2}(\sin x+\cos x)-\dfrac{e^{x}}{2}((2x-1)\sin x+\cos x)+C$

${\,I=\dfrac{e^{x}}{2}((x-1)^{2}\sin x+(x^{2}-1)\cos x)+C\,}$