How do you integrate $int x arctan x$ using integration by parts?

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Answer 1 · 2015-11-04T13:00:01

$intxtan^{-1}(x)dx=1/2[(x^2+1)tan^{-1}(x)-x]+c$ ,
where $c$ is the constant of integration.

$intxtan^{-1}(x)dx=1/2intfrac{d}{dx}(x^2)tan^{-1}(x)dx$

$=1/2[x^2tan^{-1}(x)-intx^2frac{d}{dx}(tan^{-1}(x))dx]$

$=1/2x^2tan^{-1}(x)-1/2intx^2(frac{1}{1+x^2})dx$

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

$=1/2x^2tan^{-1}(x)-1/2(x-tan^{-1}(x))+c$ ,
where $c$ is the constant of integration

$=1/2[(x^2+1)tan^{-1}(x)-x]+c$

How do you integrate int x arctan x int x arctan x using integration by parts?