Evaluate $\displaystyle \int\dfrac{dx}{\sqrt[3]{(x+1)^2(x-1)^4}}$

Question

how could I evaluate this integral

$$\int\dfrac{1dx}{\sqrt[3]{(x+1)^2(x-1)^4}}$$

Using this substitution?

$$\sqrt[n]{\frac{ax+b}{cx+d}}=t$$

substituting $\sqrt[3]{x+1}=t$ I got to

$$3\int\frac{1dt}{t^2\sqrt[3]{(t^3+2)^2}}$$

Answer 1

Observe that $$(x+1)^2(x-1)^4 = (x+1)^3(x-1)^3\cdot \frac{x-1}{x+1}$$

So you can make the change: $\frac{x-1}{x+1}=t^3$.