How do you evaluate the integral int x^2arctanx?

Jan 11, 2017

The answer is $= {x}^{3} / 3 \arctan x - \textcolor{b l u e}{\frac{1}{6} \left(- \ln \left(1 + {x}^{2}\right) + \left(1 + {x}^{2}\right)\right)} + C$

Explanation:

${\sin}^{2} x + {\cos}^{2} x = 1$, $\implies$, ${\tan}^{2} x + 1 = {\sec}^{2} x$

If $y = \arctan x$

$\tan y = x$

$\left(\tan y\right) ' = x '$

${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$, $\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 y = \frac{1}{1 + {x}^{2}}$
We use integration by parts

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

$u = \arctan x$, $\implies$, $u ' = \frac{1}{1 + {x}^{2}}$

$v ' = {x}^{2}$, $\implies$, $v = {x}^{3} / 3$

Therefore,

$\int {x}^{2} \arctan x \mathrm{dx} = {x}^{3} / 3 \arctan x - \textcolor{b l u e}{\frac{1}{3} \int \frac{{x}^{3} \mathrm{dx}}{1 + {x}^{2}}}$

Let, u=1+x^2$,$=>$,$du=2xdx#

${x}^{2} = u - 1$

so,

$\textcolor{b l u e}{\frac{1}{3} \int \frac{{x}^{3} \mathrm{dx}}{1 + {x}^{2}}} = \textcolor{b l u e}{\frac{1}{6} \cdot \int \frac{u - 1}{u} \mathrm{du}}$

$= \textcolor{b l u e}{\frac{1}{6} \left(\int - \frac{\mathrm{du}}{u} + \int \mathrm{du}\right)}$

$= \textcolor{b l u e}{\frac{1}{6} \left(- \ln u + u\right)}$

$= \textcolor{b l u e}{\frac{1}{6} \left(- \ln \left(1 + {x}^{2}\right) + \left(1 + {x}^{2}\right)\right)}$

Finally, we have

$\int {x}^{2} \arctan x \mathrm{dx} = {x}^{3} / 3 \arctan x - \textcolor{b l u e}{\frac{1}{6} \left(- \ln \left(1 + {x}^{2}\right) + \left(1 + {x}^{2}\right)\right)} + C$

Jan 11, 2017

${x}^{3} / 3 a r c \tan x - {x}^{2} / 6 + \frac{1}{6} \ln \left({x}^{2} + 1\right) + C$.

Explanation:

Let $I = \int {x}^{2} a r c \tan x \mathrm{dx}$

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

$I B P : \int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left[\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right] \mathrm{dx} .$

We take $u = a r c \tan x \therefore \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{{x}^{2} + 1}$, and,

$v = {x}^{2} \therefore \int v \mathrm{dx} = {x}^{3} / 3$

Hence, $I = {x}^{3} / 3 a r c \tan x - \frac{1}{3} \int {x}^{3} / \left({x}^{2} + 1\right) \mathrm{dx} = {x}^{3} / 3 a r c \tan x - \frac{1}{3} J ,$

where, $J = \int {x}^{3} / \left({x}^{2} + 1\right) \mathrm{dx} = \int \frac{{x}^{3} + x - x}{{x}^{2} + 1} \mathrm{dx}$

$= \int \frac{x \left({x}^{2} + 1\right)}{{x}^{2} + 1} \mathrm{dx} - \frac{1}{2} \int \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

$= \int x \mathrm{dx} - \frac{1}{2} \int \frac{\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)}{{x}^{2} + 1} \mathrm{dx}$

$\therefore J = {x}^{2} / 2 - \frac{1}{2} \ln \left({x}^{2} + 1\right)$

Altogether, $I = {x}^{3} / 3 a r c \tan x - \frac{1}{3} \left\{{x}^{2} / 2 - \frac{1}{2} \ln \left({x}^{2} + 1\right)\right\} , \mathmr{and} ,$

$I = {x}^{3} / 3 a r c \tan x - {x}^{2} / 6 + \frac{1}{6} \ln \left({x}^{2} + 1\right) + C$.

The Later Integral of J has been directly obtained using the

$\text{ Result : } \int \frac{f ' \left(x\right)}{f} \left(x\right) \mathrm{dx} = \ln | f \left(x\right) | + c$, which can easily be

proved by the substitution $f \left(x\right) = t$.

Enjoy Maths.!