Question

Integrate $\int x{\tan }^{-1}\left(x\right)\mathrm{d}x$.

Answer 1

Evaluate the given integral

Given, $\int x{\tan }^{-1}\left(x\right)\mathrm{d}x$

Let, $I=\int x{\tan }^{-1}\left(x\right)\mathrm{d}x$

Let, $t={\tan }^{-1}\left(x\right)$
$\Rightarrow \tan \left(t\right)=x$

Differentiating both sides,

$\begin{array}{cc}& \frac{\mathrm{d}x}{\mathrm{d}t}={\sec }^2\left(t\right)\\ \Rightarrow & \mathrm{d}x={\sec }^2\left(t\right)\mathrm{d}t\\ \Rightarrow & I=\int \tan \left(t\right)·t·{\sec }^2\left(t\right)\mathrm{d}t\end{array}$

From integration by parts, we know that,

$\int f\left(x\right)·g\left(x\right)=f\left(x\right)\int g\left(x\right)\mathrm{d}x-\int \left[f\text{'}\left(x\right)\int g\left(x\right)\mathrm{d}x\right]$

Here, $f\left(x\right)=t,g\left(x\right)=\tan \left(t\right){\sec }^2\left(t\right)$

$\Rightarrow I=t\int \tan \left(t\right)·{\sec }^2\left(t\right)\mathrm{d}t\mathit{-}\mathit{\int }\mathit{[}\frac{\mathit{d}\mathit{t}}{\mathit{d}\mathit{t}}\mathit{\int }\mathit{t}\mathit{a}\mathit{n}\left(t\right)\mathit{·}\mathit{s}\mathit{e}{\mathit{c}}^{\mathit{2}}\left(t\right)\mathit{d}\mathit{t}\mathit{]}$

But,

$$\begin{gathered} \int \tan \left(t\right)·{\sec }^2\left(t\right)\mathrm{d}t=\int \tan \left(t\right)·d\tan \left(t\right)\left[\because \frac{d}{dt}\tan \left(t\right)={\sec }^2\left(t\right)\right] \\ \Rightarrow \int \tan \left(t\right)·{\sec }^2\left(t\right)\mathrm{d}t=\frac{{\tan }^2\left(t\right)}{2} \end{gathered}$$

$\begin{array}{ccc}\Rightarrow & I=t·\frac{{\tan }^2\left(t\right)}{2}-\int \frac{{\tan }^2\left(t\right)}{2}\mathrm{d}t& \\ \Rightarrow & I=t·\frac{{\tan }^2\left(t\right)}{2}-\int \frac{\left[{\sec }^2\left(\mathrm{t}\right)-1\right]}{2}\mathrm{d}t& \left[\because 1+{\tan }^2\left(t\right)={\sec }^2\left(t\right)\right]\\ \Rightarrow & I=t·\frac{{\tan }^2\left(t\right)}{2}-\frac{\tan \left(t\right)}{2}+\frac{t}{2}+C& \left[\because \int {\sec }^2\left(t\right)=\tan \left(t\right)\right]\end{array}$

Replacing back $t={\tan }^{-1}\left(x\right)$ and $\tan \left(t\right)=x$, we get,

$I=\frac{x^2{\tan }^{-1}\left(x\right)}{2}-\frac{x}{2}+\frac{{\tan }^{-1}\left(x\right)}{2}+C$

Therefore, $\int x{\tan }^{-1}\left(x\right)\mathrm{d}x=\frac{x^2{\tan }^{-1}\left(x\right)}{2}-\frac{x}{2}+\frac{{\tan }^{-1}\left(x\right)}{2}+C$.