Question

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

Answer 1

`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`