Question

$\int e^{{\tan }^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)dx$.

Answer 1

$\int e^{ta{n}^{-1}x}\left(\frac{1+x+x^2}{1+x^2}\right)dx$

put $ta{n}^{-1}x=t$

x= tan t

$dx=se{c}^{2}t$ $dt$

$\int e^t\left(\frac{1+tant+ta{n}^{2}t}{1+ta{n}^{2}t}\right)se{c}^{2}tdt$

$(se{c}^{2}x-ta{n}^{2}x=1\Rightarrow 1+ta{n}^{2}x=se{c}^{2}x)$

$\int e^t\left(\frac{tant+se{c}^{2}t}{se{c}^{2}t}\right)se{c}^{2}tdt$

$\int e^t(\frac{tantse{c}^{2}t}{se{c}^{2}t}+\frac{se{c}^{2}tse{c}^{2}t}{se{c}^{2}t})dt$

$\int e^t(tant+se{c}^{2}t)dt$

$\int e^{t}tantdt+\int e^{t}se{c}^{2}tdt$

Integrate by parts $I_1$

$tant{e}^t-\int se{c}^{2}t\int e^{t}dt+\int e^{t}se{c}^{2}tdt$

$tant{e}^t-\int se{c}^{2}t\int e^{t}dt+\int e^{t}se{c}^{2}tdt$

$=e^{t}ta{n}^t+C$

Now put $t=ta{n}^{-1}x$

$=e^{ta{n}^{-1}x}tanx\left(ta{n}^{-1}x\right)+C$

$=e^{ta{n}^{-1}x}+C$

$=x{e}^{ta{n}^{-1}x}+C$