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

1 Answer
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Nov 4, 2015

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

Explanation:

#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#

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