How do you evaluate the integral #int ln(x^2+x^3)#?

1 Answer
Jan 14, 2017

#xln(x^2+x^3)+ln(abs(x-1))-3x+C#

Explanation:

We have the integral:

#I=intln(x^2+x^3)dx#

We should apply integration by parts, which takes the form #intudv=uv-intvdu#. Where #intln(x^2+x^3)dx=intudv#, let #u=ln(x^2+x^3)# and #dv=dx#.

Differentiating #u# and integrating #dv# gives:

#{(u=ln(x^2+x^3)" "=>" "du=(2x+3x^2)/(x^2+x^3)dx=(2+3x)/(x+x^2)dx),(dv=dx" "=>" "v=x):}#

So:

#I=uv-intvdu#

#I=xln(x^2+x^3)-int(2x+3x^2)/(x+x^2)dx#

#I=xln(x^2+x^3)-int(2+3x)/(1+x)dx#

Either perform long division on #(3x+2)/(x+1)# or do the following rewriting: #(3x+2)/(x+1)=(3(x+1)-1)/(x+1)=3-1/(x-1)#. So:

#I=xln(x^2+x^3)-(3intdx-intdx/(x-1))#

#I=xln(x^2+x^3)-(3x-ln(abs(x-1)))#

#I=xln(x^2+x^3)+ln(abs(x-1))-3x+C#