# Evaluate the indefinite integral. {eq}\int \frac{(\arctan x)^2}{x^2 + 1} \, \mathrm{d}x {/eq}

## Question:

Evaluate the indefinite integral.

{eq}\int \frac{(\arctan x)^2}{x^2 + 1} \, \mathrm{d}x {/eq}

## Integration by Substitution:

In calculus, the substitution method of integration is the useful process of integration that can change the form of the integrand of any complex function and makes it preferable to one that we can integrate more easily.

Some formulas that may help to evaluate the integral are as below.

{eq}\begin{align*} \int {{x^n}dx} &= \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\\ \int {\left( {\sin x} \right)dx} &= - \cos x + C\\ \int {\left( {\cos x} \right)dx} &= \sin x + C\\ \int {\left( {\sec x\tan x} \right)dx} &= \sec x + C\\ \int {\left( {{{\sec }^2}x} \right)dx} &= \tan x + C\\ \int {\dfrac{{dx}}{{1 + {x^2}}}} &= {\tan ^{ - 1}}x + C \end{align*} {/eq}

Given Data:

• The given function is: {eq}I = \int {\dfrac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^2}}}{{1 + {x^2}}}} dx {/eq}.

Apply change of variable method and assume: {eq}{\tan ^{ - 1}}x = t {/eq}, and differentiate it with respect to x.

{eq}\begin{align*} \dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x)} \right) &= \dfrac{d}{{dx}}\left( t \right)\\ \dfrac{1}{{1 + {x^2}}} &= \dfrac{{dt}}{{dx}}\\ \dfrac{{dx}}{{1 + {x^2}}} &= dt \end{align*} {/eq}

Now change the integrand by the new variable and integrate it.

{eq}\begin{align*} I &= \int {{t^2}} dt\\ &= \dfrac{{{t^{2 + 1}}}}{{2 + 1}}\\ &= \dfrac{{{t^3}}}{3} \end{align*} {/eq}

Substitute back: {eq}t = {\tan ^{ - 1}}x {/eq}.

{eq}I = \dfrac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^3}}}{3} {/eq}

Add an integral constant to the solution.

{eq}I = \dfrac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^3}}}{3} + C {/eq}

Thus, the answer is {eq}\bf{\dfrac{{{{\left( {{{\tan }^{ - 1}}x} \right)}^3}}}{3} + C} {/eq}.