Integral of x^2(x+6)^-6 without partial fractions

Question

I have tried to solve this integral $$\int\frac{x^2}{(x+6)^6}\,\mathrm dx$$ by using partial fractions.

Answer 1

Try substitution. Let $u = x + 6$, so $x = u-6$ and $du = dx$. Then we get

$$\int \frac{(u-6)^2}{u^6}du = \int \frac{u^2-12u+36}{u^6} du = \int \frac{1}{u^4} - \frac{12}{u^5} + \frac{36}{u^6} du$$

You should be able to complete the integral from there - and don't forget to plug back in for $x$!

Answer 2

HINT:

Write $x^2$ as

$$x^2=(x+6)^2-12(x+6)+36$$