Intergrate $\displaystyle \frac{1}{(2+x)^2}$ with respect to $\displaystyle x$
My attempt : $\displaystyle \frac{1}{(2+x)^2}$=$\displaystyle ln(2+x)^2$
i am having doubts if it is really so simple, i guess i am wrong and am here to clarify
No need for partial fractions (writing this as "partial fractions" would require $\displaystyle \frac{A}{x+ 2}+ \frac{B}{(x+ 2)^2}$ and you would find that A= 0, B= 1. In other words, you get back exactly what you had.)
Instead, let u= x+ 2. Then your integral is $\displaystyle \int \frac{du}{u^2}= \int u^{-2}du$