Tricky integration with odd power

Question

Integrate:

$$\int\frac{x^7}{x^{12}-1}\, dx$$

I tried to do this by putting $x^6=u$, but couldn't solve it. Please help me out.

Answer 1

Put $u = x^4$. Then $du = 4x^3 dx$ so $x^7 dx= \frac{1}{4} u \,du$ so the integral becomes $$\int \frac{x^7}{x^{12} -1 } dx = \frac 1 4 \int \frac{u}{u^3-1} du = \frac 1 4 \int \frac{u}{(u-1)(u^2+u+1)} du.$$ Now you can use partial fractions.

Answer 2

If you really have no idea what you're doing, here's an idea. We want the denominator gone, so we let $u= x^{12}-1 \implies du=12x^{11}dx$
$$\int\frac{x^7}{x^{12}-1}dx =\frac{1}{12}\int\frac{x^7}{x^{11}u}du = \frac{1}{12}\int\frac{1}{x^{4}u}du$$
We now note that our definition above gives $x^4 = (u+1)^{1/3}$ $$\frac{1}{12}\int\frac{1}{u(u+1)^{1/3}}du$$ Let $v= u+1$ $$\frac{1}{12}\int\frac{1}{(u-1)u^{1/3}}dv = 12\int\frac{1}{u^{4/3}-u^{1/3}}dv$$
Let $p=v^{1/3} \implies 3p^2\,dp = dv$ $$\frac{1}{4}\int\frac{p^2}{p^4-p}dp = \frac{1}{4}\int\frac{p}{p^3-1}dp$$
Now we can solve through partial fractions. Of course, this is WAY harder than just using the substitution $u=x^4$, but can be useful if you can't think of a good substitution. Back when I was taking early Calculus classes I would often get stressed during tests and miss easy substitutions; however, when I got stressed I just went through a process like this where I just substitute over and over again until I get something I know how to work with. Many will disagree, but I find this to be a practical thing!