would you possibly be able to show me step by step how to integrate this with respect to x:
((-x)^2)/(1+x^3)
Thanks
$\displaystyle \int \frac{(-x)^2}{1+x^3} \, dx$
$\displaystyle \int \frac{x^2}{1+x^3} \, dx$
$\displaystyle \frac13 \int \frac{3x^2}{1+x^3} \, dx$
Now make the substitution $\displaystyle u = x^3 \Rightarrow 3x^2 \, dx = du$
$\displaystyle \frac13 \int \frac{1}{1+u} \, du = \frac13\ln |u + 1| + C= \frac13\ln|1+x^3| +C$