Question
Evaluate the following integral:
$\int\frac{\text{x}^4+1}{\text{x}^2+1}\text{ dx}$

Answer

$\int\Big(\frac{\text{x}^4+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big(\frac{\text{x}^4-1+1+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big[\frac{(\text{x}^4-1)}{\text{x}^2+1}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[\frac{(\text{x}^2-1)(\text{x}^2+1)}{(\text{x}^2+1)}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[(\text{x}^2-1)+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}(\text{x})+\text{C}$

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