Why does this whole term go away with the $dx$?

Question

I'm learning how to take indefinite integrals with U-substitutions on khanacademy.org, and in one of the videos he says that: $$\int e^{x^3+x^2}(3x^2+2x) \, dx = e^{x^3+x^2} + \text{constant}$$ I understand that the differential goes away, but not how the whole $(3x^2+2x)$ term go away together with the $dx$.

Answer 1

The $u$ substitution here is $u = x^3 + x^2$. Then $du = (3x^2 + 2x) \, dx$. This allows us to rewrite

$$\int e^{x^3 + x^2}(3x^2 + 2x) \, dx = \int e^u \, du$$

At which point we integrate and substitute back. You can see that you cannot do a $u$-substitution without accounting for the change from $dx$ to $du$, and it is this that takes care of the $(3x^2 + 2x)$.

Answer 2

Using a substitution, you have

$$u=x^3+x^2,du=(3x^2+2x)dx$$

The $du$ value is the derivative of the $u$ value with respect to $x$. As part of the substitution, the $dx$ value must be changed to support integration over the substituted variable. So we have the following:

$$\int e^{x^3+x^2}(3x^2+2x)dx=\int e^udu$$

which easily evaluates to

$$e^u+C=e^{x^3+x^2}+C$$

So the $(3x^2+2x)dx$ portion is simply being replaced by $du$ as that is the value it has.