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#