Question

Find the value of $\int \frac{1}{(x^2+1)}dx$ in terms of an inverse trignometric function.

Answer 1

Let $x=tan\theta$

$\Rightarrow dx=se{c}^2\theta d\theta$

$\Rightarrow \int \frac{1}{ta{n}^2\theta +1}\times se{c}^2\theta d\theta$

$\Rightarrow \int \frac{se{c}^2\theta }{se{c}^2\theta }d\theta$

$=\theta =ta{n}^{-1}x$

$\therefore$value of integral =$ta{n}^{-1}x$