Strategies for u-Substitutions

There are mainly two types of integrals where $u$-substitutions can be useful.

The first kind is where linear terms, $kx + b$, are present. This case is greatly discussed in the "What is $u$-substitution?" section. We typically put $u = kx + b$. Why can we use $u$-substitution here? Well, simply because the integral of a product of $f(kx + b) = f(u)$ is just $k$ times the integral of $f(x)$, thereby simplifying the integral.

The second part is demonstrated in Kyle Zigner's explanation in this section, "Strategies for $u$-substitutions", where both $f(p)$ and $f'(p)$ are present in the same integral. Then we substitute $f(p) = u$ and $f'(p) = du$. Therefore, the original integral, for example $\int (4x) x^2 \, dx$ can be written as $\int 2u du$, where $u = x^2$. Because we know what $\int 2u du$ is for general $u$, substituting $u$ back can also greatly reduce our workload. Why can we use $u$-substitution here? Well, because this can cut off a part of a product of functions and save it for later substitution.