How do you integrate #x*arctan(x) dx#?

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Answer 1 · 2016-07-22T18:22:07

#= (arctan x)/2 ( x^2 + 1) - x/2 + C#

we can use IBP

#int x arctan(x) \ dx = int \ d/dx(x^2/2) arctan x \ dx#

#= (x^2/2) arctan x - int \ x^2/2 d/dx( arctan x) \ dx#

#= (x^2/2) arctan x - int \ x^2/2 * 1/(x^2 +1) \ dx#

#= (x^2/2) arctan x - 1/2 int \ x^2/(x^2 + 1) \ dx#

#= (x^2/2) arctan x - 1/2 int \ (x^2+1-1)/(x^2 + 1) \ dx#

#= (x^2/2) arctan x - 1/2 int \ 1 -1/(x^2 + 1) \ dx#

#= (x^2/2) arctan x - 1/2 (x - arctan(x)) + C#

#= (arctan x)/2 ( x^2 + 1) - x/2 + C#