Question

Fill in the blanks to complete each of the following integration formulas.

$\int {\tan }^{-1}xdx=\_\_\_$

Short answer

The result is$x\arctan \left(x\right)-\frac{1}{2}\ln \left|x^2+1\right|+C$.

Step-by-step solution

Step 1. Given information.

Consider the given integration.

$\int {\tan }^{-1}xdx$

Step 2. Find the integration.

Apply the by-parts integration method. Consider the values as $u=1$and $v={\tan }^{-1}x$.

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