Solution to the general integral $\int {f'(x)\over a^2 - [f(x)]^2} dx$?

Question

I am covering the various methods of integration in my uni course (BSc Maths). Among these, I have come up with few general results starting from integrals whose form is similar to the one above (but different signs).

Now, the problem with this one is that I am not sure how to handle it. I am pretty sure that I am to use the substitution $x=\tanh(u)$, however, I have no idea of how to carry out a general result.

Thanks in advance for your precious help.

Answer 1

Substitute $f(x)=y$, from where you have $f'(x)\mathrm dx=\mathrm dy$, and then apply partial fraction decomposition as $$\int\frac{f'(x)}{a^2-f^2(x)}\mathrm dx\stackrel{f(x)=y}=\int\frac{\mathrm dy}{a^2-y^2}=\frac1{2a}\int\frac1{a+y}+\frac1{a-y}\mathrm dy$$ These integrals are elementary and using the logarithmic representation of the $\operatorname{artanh}$ function you will arrive at $\frac1a\operatorname{artanh}\left(\frac{f(x)}a\right)+C$ as general anti-derivative of your integral. I will leave the details to you.

Answer 2

HINT

By $u=f(x) \implies du=f'(x)dx$

$$\int {f'(x)\over a^2 - [f(x)]^2} dx=\int {1\over a^2 - u^2} du=\frac1{2a}\int {1\over a - u} du+\frac1{2a}\int {1\over a + u} du$$