Question

Integrate the function $x {\tan}^{-1}x$.

Answer 1

Hint: We know that this function is not integrable directly. So, we will use integration by parts to solve this function. The formula is given by-
$\int {\text{v}}\dfrac{{du}}{{dx}}.dx = vu - \int {\text{u}}\dfrac{{dv}}{{dx}}dx$

Complete step-by-step solution -
We don’t know the integration of tan-1x, so we will assume v = tan-1x and u = x. So, we can write that-
$\Rightarrow \int x{\tan ^{ - 1}}xdx = \left[ {{{\tan }^{ - 1}}x\int xdx} \right] - \int \left[ {\left( {\int xdx} \right).\dfrac{{d{{\tan }^{ - 1}}x}}{{dx}}} \right]dx$
$\Rightarrow \int x{\tan ^{ - 1}}xdx = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \int \dfrac{{{x^2}}}{{2\left( {1 + {x^2}} \right)}}dx$
$\Rightarrow \int x{\tan ^{ - 1}}xdx = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int \dfrac{{{x^2} + 1 - 1}}{{\left( {1 + {x^2}} \right)}}dx$
$\Rightarrow \Rightarrow \int x{\tan ^{ - 1}}xdx = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int 1 - \dfrac{1}{{1 + {x^2}}}dx$
$\Rightarrow \int x{\tan ^{ - 1}}xdx = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\left( {x - {{\tan }^{ - 1}}x} \right) + c$
$\Rightarrow \int x{\tan ^{ - 1}}xdx = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}x + \dfrac{1}{2}{\tan ^{ - 1}}x + c$
This is the required answer.

Note: We should choose wisely which part we should choose as u and v. This is because we have to integrate one of the two functions. So choose carefully which function to integrate. Also, we should always add the constant of integration when we solve indefinite integration.