Solving integral with substitution gives wrong result I want to solve the integral:
$$ \int_{-\infty}^{\infty} \frac{x^2}{(1+|x|^3)^4} \ dx.$$
I thought I could solve that via the substitution $u=1+|x|^3$ since then one has the integral:
$$ \int_{-\infty}^{\infty} \frac{1}{3u^4} \ du.$$
But that seems to be wrong (probably the substitution does not work here since it is not differentiable at 0), and that's why this integral is 0, but it should be $\frac29$. So how do I solve this correctly?

Since $x^2dx=1/3d(x^3)$ using $u=x^3$ we have
$$\int_0^\infty \dfrac{x^2}{(1+x^3)^4}=\int_0^\infty1/3\dfrac 1{(1+u)^4}du=1/3\left.\left( -1/3 \dfrac1{(1+u)^3}\right)\right|=1/9.$$
Now in your integral the function is even, so the answer is just $2\cdot 1/9=2/9$
