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?
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