Question

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

Answer 1

Apply Integration by parts:

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

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

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

$\Rightarrow x=\tan \left(t\right)$

Differentiating both sides with respect to $t$,

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

From integration by parts, we know that

$\mathbf{\int }\mathbf{f}\left(x\right)\mathbf{·}\mathbf{g}\left(x\right)\mathbf{=}\mathbf{f}\left(x\right)\mathbf{\int }\mathbf{g}\left(x\right)\mathbf{dx}\mathbf{-}\mathbf{\int }\left(\frac{df\left(x\right)}{dx}\int g\left(x\right)\mathrm{dx}\right)\mathbf{dx}$

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

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

Substituting back the value of $t$,

$\begin{array}{ccc}\Rightarrow & I={\tan }^{-1}\left(x\right)·x-\ln \left(\left|\sec \left({\tan }^{-1}\left(\mathrm{x}\right)\right)\right|\right)+\mathrm{C}& \\ \Rightarrow & I=x{\tan }^{-1}\left(x\right)-\ln \left(\sqrt{1+x^2}\right)+\mathrm{C}& \left[\because \sec \left({\tan }^{-1}\left(x\right)\right)=\sqrt{1+x^2},\left|\sqrt{1+x^2}\right|=\sqrt{1+x^2}\right]\end{array}$

Therefore, $\mathbf{\int }\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathit{d}\mathit{x}\mathbf{=}\mathit{x}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(x\right)\mathbf{-}\mathit{l}\mathit{n}\left(\sqrt{1+x^2}\right)\mathbf{+}\mathit{C}$