Let $\text{I}=\int\frac{\text{x}^2}{1+\text{x}^2}\text{ dx}$
Let $1+\text{x}^3=\text{t}$
$3\text{x}^2\text{dx}=\text{dt}$
$\text{x}^2\text{dx}=\frac{1}{3}\text{dt}$
$\therefore\ \frac{1}{3}\int\frac{\text{dt}}{\text{t}}=\frac{1}{3}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{3}\log|1+\text{x}^3|+\text{C}$
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