Question

Solve $\int e^x\left\{\frac{1+(1+x^2){\tan }^{-1}x}{1+x^2}\right\}dx.........+c(x>0)$

• A. $x{\tan }^{-1}x$
• B. $\frac{x}{1+x^2}$
• C. $e^x{\tan }^{-1}x$
• D. $\frac{e^x}{1+x^2}$

Answer 1

Answer: (C)

$e^x{\tan }^{-1}x$

$I=\int e^x\left(\frac{1+(1+x^2){\tan }^{-1}x}{1+x^2}\right)dx$
$=\int e^x(\frac{1}{1+x^2}+{\tan }^{-1}x)dx$
$=\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx$
take $f\left(x\right)={\tan }^{-1}x$
$\therefore f^{\prime }\left(x\right)=\frac{1}{1+x^2}$
$\therefore I=e^{x}f\left(x\right)+c$
$\therefore I=e^x{\tan }^{-1}x+c$