Question

What is the integral of `int ((x^2-1)/sqrt(2x-1) )dx`?

Answer 1

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

Explanation:

Our big problem in this integral is the root, so we want to get rid of it. We can do this by introducing a substitution `u=sqrt(2x-1)`. The derivative is then
`(du)/dx=1/sqrt(2x-1)`

So we divide through (and remember, dividing by a reciprocal is the same as multiplying by just the denominator) to integrate with respect to `u`:
`int\ (x^2-1)/sqrt(2x-1)\ dx=int\ (x^2-1)/cancel(sqrt(2x-1))cancel(sqrt(2x-1))\ du=int\ x^2-1\ du`

Now all we need to do is express the `x^2` in terms of `u` (since you can't integrate `x` with respect to `u`):
`u=sqrt(2x-1)`

`u^2=2x-1`

`u^2+1=2x`

`(u^2+1)/2=x`
`x^2=((u^2+1)/2)^2=(u^2+1)^2/4=(u^4+2u^2+1)/4`

We can plug this back into our integral to get:
`int\ (u^4+2u^2+1)/4-1\ du`

This can be evaluated using the reverse power rule:
`1/4*u^5/5+2/4*u^3/3+u/4-u+C`

Resubstituting for `u=sqrt(2x-1)`, we get:
`1/20(2x-1)^(5/2)+1/6(2x-1)^(3/2)-3/4sqrt(2x-1)+C`