∫(x^2)e^[-(x^3)/4]

Juz confusing that we didnt take **(x^2)** into account when integrating? Juz need to integrate **e^[-(x^3)/4] **so it becomes:

**[x^2] * [-4/3(x^2)]e^[-(x^3)/4]**

and so the **(x^2) **just cancel out itself to become **(-4/3)e^[-(x^3)/4]**

You actually do take the \(\displaystyle x^2\) term into account!

When you make the substitution \(\displaystyle u={\color{blue}-\tfrac{1}{4}x^3}\), we see that \(\displaystyle \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}\)

At this stage, let's go back to the original integral:

\(\displaystyle \int {\color{red}x^2}e^{-x^3/4}{\color{red}\,dx}\).

Observed that I highlighted the part that appeared in our \(\displaystyle \,du\) term! So we see that \(\displaystyle \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}\implies -\tfrac{4}{3}\,du={\color{red}x^2\,dx}\).

Therefore, \(\displaystyle \int {\color{red}x^2}e^{{\color{blue}-x^3/4}}{\color{red}\,dx}=\int e^u(-\tfrac{4}{3}\,du)=-\tfrac{4}{3}\int e^u\,du\).

Then integrating results in \(\displaystyle -\tfrac{4}{3}e^u+C\) and back substitution gives us the desired result \(\displaystyle \int x^2e^{-x^3/4}\,dx=-\tfrac{4}{3}e^{-x^3/4}+C\).

Does this clarify what's going on here?