Question

How do you solve this integral?

`intx^3(x^2-4)^(3/2)`

Answer 1

`I=1/7(x^2-4)^(7/2)+4/5(x^2-4)^(5/2)+C`

Explanation:

We want to solve

`I=intx^3(x^2-4)^(3/2)dx`

Use substitution let `u=x^2-4=>(du)/dx=2x`

`I=intx^3(u)^(3/2)1/(2x)du`

`=1/2intx^2(u)^(3/2)du`

But `u=x^2-4<=>x^2=u+4`

`I=1/2int(u+4)(u)^(3/2)du`

`=1/2intu(u)^(3/2)du+1/2int4(u)^(3/2)du`

`=1/2intu^(5/2)du+2intu^(3/2)du`

By the power rule for the integrals

`I=1/7u^(7/2)+4/5u^(5/2)+C`

Substitute `u=x^2-4` back

`I=1/7(x^2-4)^(7/2)+4/5(x^2-4)^(5/2)+C`