# Proving integration formula involving the form a+bx

While trying to memorize and understand various integration formulas, I came across an integration rule stating that $$\int \frac{1}{x^2(a+bx)^2} dx = -\frac{1}{a^2}\left[\frac{a+2bx}{x(a+bx)}+\frac{2b}{a} \ln{ \left|\frac{x}{a+bx} \right| }\right] +C$$

I am stuck on how to prove the formula, save for the obvious fact that we can differentiate the right side and seeing that it works. I used various integration calculators (with steps) online, but I still cannot figure out how that integration formula works. If this is of any help, I already understand that

$$\int \frac{1}{x(a+bx)} du = \frac{1}{a} \ln{\left|\frac{x}{a+bx} \right|}+C$$ and that $$\int \frac{1}{x^2(a+bx)} du = -\frac{1}{a}\left[\frac{1}{x}+\frac{b}{a} \ln{\left|\frac{x}{a+bx} \right|}\right]+C$$ and that $$\int \frac{1}{x(a+bx)^2} du = \frac{1}{a}\left[\frac{1}{a+bx}+\frac{1}{a} \ln{\left|\frac{x}{a+bx} \right|}\right]+C$$ The reason that I posted the latter three here is that I suspect that we might be able to simplify the first integral to one of the last three integrals.

• Note: to make the brackets bigger use the syntax \left[ and \right] . If you click on "edit" you can see the changes I made to your formatting.
– lulu
May 21 '16 at 1:28
• PLease, change $du$ to $dx$. May 21 '16 at 4:36
• You mean differentiate the right side May 21 '16 at 6:42

For all cases, I think that the easiest is to start using partial fraction decomposition. This would give $$\frac{1}{x(a+bx)}= \frac{1}{a x}-\frac{b}{a (a+b x)}$$ $$\frac{1}{x^2(a+bx)}= \frac{b^2}{a^2 (a+b x)}-\frac{b}{a^2 x}+\frac{1}{a x^2}$$ $$\frac{1}{x(a+bx)^2}=-\frac{b}{a^2 (a+b x)}+\frac{1}{a^2 x}-\frac{b}{a (a+b x)^2}$$ $$\frac{1}{x^2(a+bx)^2}=\frac{2 b^2}{a^3 (a+b x)}-\frac{2 b}{a^3 x}+\frac{b^2}{a^2 (a+b x)^2}+\frac{1}{a^2 x^2}$$ At this point, all integrals are now simple.