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.

  • $\begingroup$ 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. $\endgroup$
    – lulu
    May 21 '16 at 1:28
  • $\begingroup$ PLease, change $du$ to $dx$. $\endgroup$ May 21 '16 at 4:36
  • $\begingroup$ You mean differentiate the right side $\endgroup$
    – Nikunj
    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.

For the last one, after integration and simplifications, you will get the result.


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