Answer
$\frac{\ln(2)}{6}$
Work Step by Step
The x value of the 2 intersection points are 0 and 1, the lower and upper bounds.
$\int\frac{x}{1+x^2}=\frac{\ln(x^2+1)}{2}$
$\int\frac{x^2}{1+x^3}=\frac{\ln|x^3+1|}{3}$
$\int_0^1\Big(\frac{x}{1+x^2}-\frac{x^2}{1+x^3}\Big)dx=\Big[\frac{\ln(x^2+1)}{2}-\frac{\ln|x^3+1|}{3}\Big]_0^1=\frac{\ln(2)}{6}$
