How to solve: $\int \frac{1}{36x^2(1+12x)^2}\mathrm dx$ I was given the following problem: $$\displaystyle\int \frac{1}{36x^2(1+12x)^2}\mathrm dx$$ I was told to solve it using the following formula from an integral table: $$\int\frac{\mathrm du}{u^2(a+bu)^2}=-\frac1{a^2}\left[\frac{a+2bu}{u(a+bu)}+\frac{2b}a\ln\left|\frac{u}{a+bu}\right|\right]+C$$
I am not sure how to plug this in. I do see that you can pull $\frac1{36}$ out of the problem, and that we have $x^2$ and $(1+12x)^2$ in the denominator, but what do we need to set as our variables?
Let $u=6x$. Then $u^2 = 36x^2$, $2u=12x$, and $du=6 \;dx$. Your integral becomes

$$\int \frac{1}{36x^2(1+12x)^2}dx = \int \frac{1}{u^2(1+2u)^2}\cdot \frac 1 6 \; du$$

You can solve it from here using your formula.
Hint

Express the numerator as $$(12x+1-12x)^2=(12x+1)^2+(12x)^2-24x(12x+1)$$

Again $$\dfrac1{x(12x+1)}=\dfrac{12x+1-12x}{x(12x+1)}=?$$