How do I solve the integral: $\int x^2\ln(x^6-1)dx$ using integration by parts?

Question

As the title says I need to solve: $\int x^2\ln(x^6-1)dx$.
Now I said that $$\int x^2\ln(x^6-1)dx=\frac{x^3\ln(x^6-1)}{3}-\int \frac{6x^8}{3(x^6-1)}dx=\frac{x^3\ln(x^6-1)}{3}-2\int \frac{x^8}{(x^6-1)}dx$$.
The problem is I do not see any way to solve:$\int \frac{x^8}{(x^6-1)}dx$.
Please help.

Answer 1

You can start by factoring the expression in your logarithm and turn this into

$$\int{x^2 \ln(x^3+1)}\, dx + \int{x^2 \ln(x^3-1)}\, dx.$$

Both of these are integrals of the form $\int \ln u\, du$, which can be solved by integration by parts.

Answer 2

Make the substitution $u=x^3\implies\mathrm{d}u=3x^2\mathrm{d}x$ this changes the integral into $\frac{1}{3}\int\frac{u^2}{u^2-1}\,\mathrm{d}u$ which can be solved by long division then partial fractions