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Q13  Integrate the functions \tan ^{-1} x

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Consider \tan ^{-1} x

So, we have then: I =\int 1.\tan^{-1}x dx

After taking \tan^{-1}x as a first function and 1 as second function and integrating by parts, we get

I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx

= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx

= x\tan^{-1}x -\frac{1}{2}\int \frac{2x}{1+x^2}dx

= x\tan^{-1}x -\frac{1}{2}\log|1+x^2|+C

= x\tan^{-1}x -\frac{1}{2}\log(1+x^2)+C

Posted by

Divya Prakash Singh

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