My integration skills leave a lot to be desired, but noone has attempted an answer yet, so I will have a go. At least someone can then pick holes in my poor attempts.

First of all, shouldn't that be d²y/dx²=x/(a+x) ?

Now, ∫ 1/x dx = ln x , and ∫ 1/f(x) dx = ln ( f(x) ) , so:

∫ x/(a+x) dx = ∫ 1 - a/(a+x) dx = ∫ dx - a ∫ 1/(a+x) dx = x - a ln(a+x) + C

Now I managed to find that ∫ ln(a+x) dx = a ln x + ½ ln x²

∫ x-a ln(a+x) + C dx = ∫ x dx - a ∫ ln(a+x) dx + C ∫ dx= x²/2 - a( a ln x - ½ ln x²) + Cx + D

Can someone please check my work?