# Q. `int` `(x^6 - 1)/(x^2 + 1)` `dx` = ?

### 1 Answer | Add Yours

Using long division it can be obtained that;

`(x^6-1)/(x^2+1) = (x^4-x^2+1)-2/(x^2+1)`

`int(x^6-1)/(x^2+1)dx `

`= int((x^4-x^2+1)-2/(x^2+1))dx`

`= int(x^4-x^2+1)dx-2int(1/(x^2+1))dx`

`= x^4/4-x^3/3+x-2tan^(-1)x+C` where C is constant

*So the answer is*

`int(x^6-1)/(x^2+1)dx` = `x^4/4-x^3/3+x-2tan^(-1)x+C`

**Sources:**