Integrate x^3/2 divided by expression - using partial fractions perhaps

[Mustaq]

Homework Statement

Hi. My first post!
I'm trying to solve for where a is a constant:

∫ (x/a)^1/2*(x/(x-a)) dx

Homework Equations

See above

The Attempt at a Solution

I've tried integration by parts by setting u=(x/a)^1/2 but I end up having to solve ∫ (x/a)^1/2ln(x-a) - which I can't solve.
I've tried switching them around u=x/(x-a) and end up having to solve ∫x^3/2/(x-a)² - which I can't solve either.
I've thought of using partial fractions but run into x^3/2/(x-a)² which I can't do either.

Any help gratefully received.
Thanks

 

[sunjin09]

The change of variable u=(x/a)^1/2 is very much solvable. Try it again.

 

[Mustaq]

So obvious. I was looking for something really complicated.
Let u = (x/a)^1/2
So x=au², dx/du=2au

First simplify:
∫(x/a)^1/2(x/(x-a)) dx
∫(x/a)^1/2( 1 + a/(x-a) ) dx

Substitute by u:
∫u( 1 + a/(au²-a) ) 2au du
2a∫u²( 1 + 1/(u²-1) ) du
2/3au³ + 2a∫u²/(u²-1) du
2/3au³ + 2a∫(1 + 1/(u²-1)) du
2/3au³ + 2au + aln( (u-1)/(u+1) )

Finally put back x and get required anaswer:
2/3(x³/a)^1/2 + 2(ax)^1/2 + aln( ((x/a)^1/2-1)/((x/a)^1/2+1) )

Thanks, for the hint which lead to the solution (was on it for weeks LOL)