Thread: Integrating

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

Have you seen partial fractions before?

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 $\displaystyle \int \frac{du}{u^2}= \int u^{-2}du$

does this means i do not use ln rule to integrate but rather use normal power rule to intergrate?

Originally Posted by Punch

does this means i do not use ln rule to integrate but rather use normal power rule to intergrate?

No, you do not need the ln rule. Yes, you need the power rule.

$\displaystyle \int \frac{1}{x} dx = ln(x)$ does NOT mean that $\displaystyle \int \frac{1}{x^2} dx = ln(x^2)$

Follow what HallsofIvy has said in the above post.