# Find this integration

Gold Member

## Homework Statement

$\displaystyle \int_0^1 \dfrac{xe^{tan^{-1}x}}{\sqrt{1+x^2}} dx$

## The Attempt at a Solution

Let tan^-1 (x) = t
x = tant
dt=dx/sqrt{1+x^2}
The integral then reduces to

$\displaystyle \int_0^{\pi/4} tante^tdt$

Applying integration by parts by taking tant as 1st function

$tant e^t - \displaystyle \int sec^2te^t dt$

This has made the problem more complicated instead of simplifying.

## Homework Statement

$\displaystyle \int_0^1 \dfrac{xe^{tan^{-1}x}}{\sqrt{1+x^2}} dx$

## The Attempt at a Solution

Let tan^-1 (x) = t
x = tant
dt=dx/sqrt{1+x^2}
The integral then reduces to

$\displaystyle \int_0^{\pi/4} tante^tdt$

Applying integration by parts by taking tant as 1st function

$tant e^t - \displaystyle \int sec^2te^t dt$

This has made the problem more complicated instead of simplifying.

$$\frac{d}{dx}\left(\arctan(x)\right) ≠ \frac{1}{\sqrt{1+x^2}}$$
To use the substitution, multiply and divide by ##\sqrt{1+x^2}##.