Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{1-x^4}{1-x}\right)dx \\ =\int \frac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x\right)}dx \\ =\int \frac{\left(1-x\right)\left(1+x\right)\left(1+x^2\right)}{\left(1-x\right)}dx \\ =\int \left(1+x\right)\left(1+x^2\right)dx \\ =\int \left(1+x^2+x+x^3\right)dx \\ =x+\frac{x^3}{3}+\frac{x^2}{2}+\frac{x^4}{4}+C \end{gathered}$$