How do you integrate #int xln(1+x)# using integration by parts?

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Answer 1 · 2016-10-22T09:12:34

#I=1/2x^2ln(1+x)-1/4x^2+1/2x-1/2ln(1+x)+C#

#intu(dv)/dxdx=uv-intv(du)/dxdx#

let #u=ln(1+x)=>(du)/dx=1/(1+x)#

#(dv)/dx=x=>v=1/2x^2#

#I=1/2x^2ln(1+x)-1/2int(x^2/(1+x))dx#

#I=1/2x^2ln(1+x)-1/2int(x-x/(1 +x))dx#

#I=1/2x^2ln(1+x)-1/2int(x-1+1/(x+1))dx#

#I=1/2x^2ln(1+x)-1/4x^2+1/2x-1/2ln(1+x)+C#