# Hint for solving an integral

How can i solve the following integral? $$\int \frac{-x f´}{(a^2 - x^2)f} dx$$ Where $f:= f(x)$ is an integrable function.

Thank you for any help.

• I don't think you can. – Yves Daoust Nov 24 '16 at 9:27
• What do you mean by "solve" ? Finding a simpler expression, in particular with $f$ present once only ? – Jean Marie Nov 24 '16 at 9:31
• Where did you get this problem from? – A.Γ. Nov 24 '16 at 9:40

If $(a^2-x^2)f(x)>0$ consider $F(x):= \ln((a^2-x^2)f(x))$ and compute $F'$

• This doesn't lead to the solution, does it ? – Yves Daoust Nov 24 '16 at 8:53
• No. You are right and I am wrong. Thanks – Fred Nov 24 '16 at 8:57
• Did you try partial fractions – hamam_Abdallah Nov 24 '16 at 9:17
• @AbdallahHammam Quite useless considering that you know nothing about $f$. Also, questions need a "?" at their end. – Von Neumann Nov 24 '16 at 9:21

Try with integration by parts, noticing that

$$\frac{f'(x)}{f(x)} = \frac{\text{d}}{\text{d}x}\ln\left(f(x)\right)$$

$$-\int \frac{x}{a^2-x^2} \frac{f'(x)}{f(x)}\ \text{d}x = -\int \underbrace{\frac{x}{a^2-x^2}}_{u(x)}\ \underbrace{\frac{\text{d}}{\text{d}x}\ln\left(f(x)\right)}_{w'(x)}\ \text{d}x$$

Getting a partial result like

$$-\left(\frac{x}{a^2-x^2}\ln(f(x) - \int \left(\frac{1}{a^2-x^2} + \frac{2x^2}{(a^2-x^2)^2}\right)\ln f(x)\right)$$

The unknown $f(x)$ does not seem to lead us too much further.