# How to evaluate this integral: $\int \frac{x^4}{(x-1)(x^2-1)}dx\;$?

$$\int \frac{x^4}{(x-1)(x^2-1)}dx$$

I tried to decompose the $(x^2-1)$ term into $(x+1)(x-1)$ thus getting $(x-1)^2(x+1)$ as the denominator. I can't use the method of partial fraction because of the $x^4$ term. Should I proceed through normal polynomial(I'm finding it difficult) division or is there any other methods using trigonometric substitution to solve this. Please provide only hints so that I can work it out myself.

• I am Quite Happy for "Please provide only hints so that I can work it out myself." – user87543 Jan 9 '14 at 14:11

## 2 Answers

Hint:

Then you can use partial fraction decomposition, where appropriate.

• Sir, I got the answer. Thank you for your hint. But, is there any other elegant way to solve this? – Rajath Krishna R Jan 9 '14 at 14:30
• @RajathKrishnaR it is Ma'am. am-y... – Lost1 Jan 9 '14 at 22:41

Hint:

$$\frac{x^4}{(x-1)(x^2-1)}=\frac{x^4-1+1}{(x-1)(x^2-1)}=\frac{x^2+1}{x-1}+\frac{1}{(x-1)(x^2-1)}$$

$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$$

$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)^2}-\frac{1}{4(x-1)}+\frac{1}{4(x+1)}$$

$$\to \int \frac{x^4}{(x-1)(x^2-1)}dx=\frac{x^2}{2}+x-\frac{1}{2(x-1)}+\frac{7}{4}\ln(x-1)+\frac{1}{4}\ln(x+1)+\text{C}$$