How do you solve this integral?

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

1 Answer
Feb 16, 2018

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