MHB Integral: Solving the Difficult One

[Natarajan]

Consider the following:

$$\int \left(\frac{x-1}{x+1}\right)^4\,dx$$

I am unable to solve this.

 

[MarkFL]

Hello and welcome to MHB, Natarajan!

I have moved your thread since this forum is a better fit for your question.

I think the first thing I would do is write:

$$\frac{x-1}{x+1}=\frac{x+1-2}{x+1}=1-\frac{2}{x+1}$$

Let's substitute:

$$u=x+1\implies du=dx$$

Now, apply the binomial theorem:

$$\left(1-2u^{-1}\right)^4=16u^{-4}-32u^{-3}+24u^{-2}-8u^{-1}+1$$

Now you can integrate term by term, and then back-substitute for $u$. Can you proceed?