Q:

Integration method unknown

$$
\int \frac{4}{(x+y)^{2}} \,\mathrm{d}y = \frac{-4}{x+y}
$$
How do you get to that answer? I've been looking over my formulas, tried to compute the power and integrate after but no success.

A:

$$
\int \frac{4}{(x+y)^{2}} \,\mathrm{d}y = \frac{-4}{x+y}
$$

Factor out constant:
$$
4\int \frac{1}{(x+y)^{2}}dx \
$$
For integrand $\frac{1}{(x+y)^2}$ substitute $u=x+y$ and $du=dy$:
$$
4\int \frac{1}{u^{2}}du \
$$
The integral of $\frac{1}{u^2}$ is $=-\frac{1}{u}$
$$
=-\frac{4}{u}+C
$$
Substitute back for $u=x+y$
$$
\color{blue}{=-\frac{4}{x+y}+C}
$$

