Need help simplifying this fraction for integration

Question

$$\int \frac{x^6 - 1}{1 + x^2}\ dx\ =\ ?$$

I know that I have to do something like divide the fraction, but I am not getting how, Can someone please help me simplify this?

Answer 1

Divide top by bottom: $$\int \frac{x^6 - 1}{1 + x^2}\ dx=\int \left(x^4-x^2+1-\frac{2}{x^2+1}\right)\,dx$$

$$= \int \left(x^4-x^2 + 1\right)\,dx - 2\int \frac 1{1+x^2}\,dx$$

The first three terms in the sum can be readily integrated using the "power rule".

For the last term, let $x = \tan^{-1}\theta$, and take it from here (inverse tangent).

Answer 2

The first thing we need to do is divide $ \frac{x^6-1}{1+x^2} $.

Doing this gives us

$$ {x^6-1} = (x^4 - x^2 + 1)(1+x^2) - 2 $$ $$ \frac{x^6-1}{1+x^2} = x^4 - x^2 + 1 - \frac{2}{1+x^2} $$

So, we can now integrate $\frac{x^6-1}{1+x^2}$

$$ \int \frac{x^6-1}{1+x^2} dx = \int (x^4 - x^2 + 1 - \frac{2}{1+x^2}) dx $$

$$ = \frac{x^5}{5} - \frac{x^3}{3} + x - 2\,arctan(x) + C $$