# How do you find the antiderivative of int (xarctanx) dx?

## How do you find the antiderivative : $\int \left(x \arctan x\right) \mathrm{dx}$?

Sep 12, 2017

$\frac{1}{2} \left\{\left({x}^{2} + 1\right) a r c \tan x - x\right\} + C .$

#### Explanation:

Suppose that, $I = \int x a r c \tan x \mathrm{dx} .$

We will use the following Method of Integration by Parts (IBP) :

IBP :$\int u \cdot v \mathrm{dx} = u \int v \mathrm{dx} - \int \left(\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right) \mathrm{dx} .$

We take, $u = a r c \tan x , \mathmr{and} , v = x .$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1} , \mathmr{and} , \int v \mathrm{dx} = {x}^{2} / 2.$

Therefore, by IBP, we have,

$I = {x}^{2} / 2 \cdot a r c \tan x - \int \left\{{x}^{2} / 2 \cdot \frac{1}{{x}^{2} + 1}\right\} \mathrm{dx} ,$

$= {x}^{2} / 2 \cdot a r c \tan x - \frac{1}{2} \int {x}^{2} / \left({x}^{2} + 1\right) \mathrm{dx} ,$

$= {x}^{2} / 2 \cdot a r c \tan x - \frac{1}{2} \int \frac{\left({x}^{2} + 1\right) - 1}{{x}^{2} + 1} \mathrm{dx} ,$

$= {x}^{2} / 2 \cdot a r c \tan x - \frac{1}{2} \int \left\{\frac{{x}^{2} + 1}{{x}^{2} + 1} - \frac{1}{{x}^{2} + 1}\right\} \mathrm{dx} ,$

$= {x}^{2} / 2 \cdot a r c \tan x - \frac{1}{2} \left\{\int 1 \mathrm{dx} - \int \frac{1}{{x}^{2} + 1} \mathrm{dx}\right\} ,$

$= {x}^{2} / 2 \cdot a r c \tan x - \frac{1}{2} \left\{x - a r c \tan x\right\} .$

$\Rightarrow I = \frac{1}{2} \left\{\left({x}^{2} + 1\right) a r c \tan x - x\right\} + C .$

Enjoy Maths.!