How do you evaluate the integral of #int x*arctan(x)#?

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Answer 1 · 2016-08-12T22:56:46

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

#int x*arctan(x) \ dx#

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

which, by IBP, is.....
#=x^2/2 *arctan(x) - int (x^2/2) d/dx arctan(x) \ dx#

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

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

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

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

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

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

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