Question

$\int \frac{x^4+1}{x^2+1}dx$

Answer 1

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