What is the integral of $int 1/(x^2+1)dx$ ?

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Answer 1 · 2018-07-07T22:46:13

$\int 1/{x^2+1}\ dx=\tan^{-1}(x)+C$

Let $x=\tan\theta\implies dx=\sec^2\theta\ d\theta$ & $\theta=\tan^{-1}(x)$

$\therefore \int 1/{x^2+1}\ dx$

$=\int 1/{tan^2\theta+1}\ \sec^2\theta\ d\theta$

$=\int \frac{\sec^2\theta\ d\thea}{\sec^2\theta}$

$=\int d\theta$

$=\theta+C$

$=\tan^{-1}(x)+C$

What is the integral of int 1/(x^2+1)dxint 1/(x^2+1)dx?