Question

Prove the integration formula.

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

(a) by applying integration by parts to $\int {\tan }^{-1}xdx$.

(a) by differentiating$x{\tan }^{-1}x-\frac{1}{2}\ln \left(x^2+1\right)$.

Short answer

Part (a). The solution is $x{\tan }^{-1}x-\frac{1}{2}\ln \left(x^2+1\right)+C$.

Part (b). The solution is${\tan }^{-1}x$.

Step-by-step solution

Part (a) Step 1. Given information.

The given integral is:

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

Part (a) Step 2. Prove the formula by applying integration by parts.

Take $u={\tan }^{-1}x$and $v=1$.

$\begin{array}{rcl}\int {\tan }^{-1}x·1dx& =& {\tan }^{-1}x\int dx-\int \left[\frac{d}{dx}\left({\tan }^{-1}x\right)\int dx\right]dx\\ & =& x·{\tan }^{-1}x-\int \frac{x}{1+x^2}dx\\ & =& x·{\tan }^{-1}x-\frac{1}{2}\int \frac{1}{t}dt\cdots \cdots \cdots \cdots \cdots \cdots \left\{put1+x^2=dt,2xdx=dt,xdx=\frac{1}{2}dt\right\}\\ & =& x·{\tan }^{-1}x-\frac{1}{2}\ln \left(t\right)+C\\ & =& x·{\tan }^{-1}x-\frac{1}{2}\ln \left(x^2+1\right)+C\end{array}$

Hence, the formula is proved.

Part (b) Step 1. Given information.

The given formula is$\int {\tan }^{-1}xdx=x·{\tan }^{-1}x-\frac{1}{2}\ln \left(x^2+1\right)+C$.

Part (b) Step 2. Prove the formula by differentiating x·tan-1x-12lnx2+1.

$\begin{array}{rcl}\frac{d}{dx}\left\{x·{\tan }^{-1}x-\frac{1}{2}\ln \left(x^2+1\right)\right\}& =& x\frac{d}{dx}\left\{{\tan }^{-1}x\right\}+{\tan }^{-1}x\frac{d}{dx}\left\{x\right\}-\frac{1}{2}\frac{d}{dx}\left\{\ln \left(x^2+1\right)\right\}\\ & =& \frac{x}{1+x^2}+{\tan }^{-1}x-\frac{1}{2}·\frac{1}{1+x^2}·2x\cdots \cdots \cdots \cdots \left\{applychainruleofdifferentiation\right\}\\ & =& \frac{x}{1+x^2}+{\tan }^{-1}x-\frac{x}{1+x^2}\\ & =& {\tan }^{-1}x\end{array}$

Hence, the formula is proved.