Solve $\int_0^x (z^2+1)e^{s\left(\frac{z^3}{3}+z\right)} \ dz$

Question

I am trying to evaluate the integral $$\int_0^x (z^2+1)e^{s\left(\frac{z^3}{3}+z\right)} \ dz.$$

I thought about using the substitution $$y=\frac{z^3}{3}+z\implies dy=z^2+1 \ dz.$$ However, I'm unsure how my bounds of integration change, given my substitution made. Do they change at all?

Answer 1

The factor $s(z^2+1)$ is the derivative of $e^{s(z^3/3+z)}$

Once you put $y=z^3 +z$, the new limits are $0$ and $x^3/3 +x$