Question

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

Answer 1

The answer is `1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C`

Explanation:

Let `u = 2x - 1`. Then `du = 2dx` and `dx = 1/2du`.

`int(x^2 - 1)/sqrt(u) * 1/2du`

We want to get rid of the x's. `x = (u + 1)/2`, so `x^2 = ((u + 1)/2)^2`

`int(((u + 1)/2)^2 - 1)/sqrt(u) * 1/2du`

Expand:

`1/2int(((u^2 + 2u + 1)/4) - 1)/sqrt(u)du`

`1/2int(u^2 + 2u + 1 - 4)/(4sqrt(u))du`

`1/8int(u^2 + 2u - 3)/sqrt(u)`

`1/8int(u^2/sqrt(u) + (2u)/sqrt(u) - 3/sqrt(u))du`

`1/8int(u^(3/2) + 2u^(1/2) - 3u^(-1/2))du`

`1/8(2/5u^(5/2) + 4/3u^(3/2) - 6u^(1/2))`

`1/20u^(5/2) + 1/6u^(3/2) - 3/4u^(1/2)`

`1/20(2x - 1)^(5/2) + 1/6(2x - 1)^(3/2) - 3/4(2x - 1)^(1/2) + C`

Hopefully this helps!

Answer 2

`int (x^2-1)/sqrt(2x-1)*dx`

=`1/4*int (4x^2-4)/sqrt(2x-1)*dx`

=`1/4*int ((2x)^2-4)/sqrt(2x-1)*dx`

After using `u=sqrt(2x-1)`, `2x=u^2+1`, `2dx=2u*du` or `dx=u*du` transforms, this integral became

`1/4*int ((2x)^2-4)/sqrt(2x-1)*dx`

=`1/4*int ((u^2+1)^2-4)/u*(u*du)`

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

=`1/4*(u^5/5+(2u^3)/3-3u)+C`

=`u^5/20+u^3/6-(3u)/4+C`

=`1/20*(2x-1)^(5/2)+1/6*(2x-1)^(3/2)-3/4*sqrt(2x-1)+C`