Integration of $\int\frac{x^{4}}{x+1}dx$ Could you help me with the following integral:
$$\int\frac{x^{4}}{x+1}dx$$

HINT: $$x^4=x^4-1+1=(x-1)(x+1)(x^2+1)+1$$
Generalization : For integer $n\ge1,$
$$x^n-1=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
For $x=y^2,y^{2n}-1=(y^2-1)(y^{2(n-1)}+y^{2(n-2)}+\cdots+y^2+1)$
$\iff y^{2n}-1=(y-1)(y+1)(y^{2n-2}+y^{2n-4}+\cdots+y^2+1)$
