Question
Integrate the function in Exercise:
$\tan^{-1}\text{x}$

Answer

Let $\text{I}=\int1.\tan^{-1}\text{x dx}$
Taking tan-1 x as first function and 1 as second function and integrating by parts, we obtain.
$\text{I}=\tan^{-1}\text{x}\int1.\text{dx}-\int\Bigg\{\Bigg(\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}\Bigg)\int1. \ \text{dx}\Bigg\}\text{dx}$
$=\tan^{-1}\text{x}.\text{x}-\int\frac{1}{1+\text{x}^2}.\text{x dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}|1+\text{x}^2|+\text{C}$
$=\text{x}\tan^{-1}\text{x}-\frac{1}{2}\text{log}(1+\text{x}^2)+\text{C}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free