Question

Integrate the function [e^(tan^{-1}x)/(1 + x²)].dx

Answer 1

HELLO DEAR,

given function is $\bold{\int{\frac{e^{tan^{-1}x}}{1 + x^2}}\,dx}$

let tan^{-1}x = t [tec]\bold{\Rightarrow 1/(1 + x^2) = dt/dx}[/tex]
$\bold{\Rightarrow dx = (1 + x^2).dt}$

now,the substitute the of t and dx in the function,we get

$\bold{\int{\frac{e^{tan^{-1}x}}{1 + x^2} * (1 + x^2)dt}}$

$\bold{\int{e^t*dt}}$

$\bold{e^t + C}$

$\bold{e^{tan^{-1}x} + C}$

where, c is arbitrary constant.

HENCE, THE INTEGRAL OF [e^(tan^{-1}x)/(1 + x²)].dx is $\bold{e^{tan^{-1}x} + C}$

I HOPE ITS HELP YOU DEAR,
THANKS