Question

How do you integrate `int x^3*dx/(x^2-1)^(2/3)`?

Answer 1

`intx^3/(x^2-1)^(2/3)dx=(3(x^2-1)^(1/3)(x^2+3))/8+C`

Explanation:

We have the integral:

`I=intx^3/(x^2-1)^(2/3)dx`

Using substitution, let `u=x^2-1`, so `du=2xdx` and `x^2=u+1`. Thus:

`I=int(x^2(x))/(x^2-1)^(2/3)dx=1/2int(x^2(2x))/(x^2-1)^(2/3)dx=1/2int(u+1)/u^(2/3)du`

Split up the fraction and then the integrals:

`I=1/2int(u^(1/3)+u^(-2/3))du=1/2intu^(1/3)du+1/2intu^(-2/3)du`

Integrate using the power rule for integrals:

`I=1/2(u^(4/3)/(4/3))+1/2(u^(1/3)/(1/3))=3/8u^(4/3)+3/2u^(1/3)=(3u^(4/3)+12u^(1/3))/8`

Continuing simplification:

`I=(3u^(1/3)(u+4))/8=(3(x^2-1)^(1/3)((x^2-1)+4))/8=(3(x^2-1)^(1/3)(x^2+3))/8+C`