# Integrating

• Jun 2nd 2010, 01:44 AM
Punch
Integrating
Intergrate $\frac{1}{(2+x)^2}$ with respect to $x$

My attempt : $\frac{1}{(2+x)^2}$= $ln(2+x)^2$

i am having doubts if it is really so simple, i guess i am wrong and am here to clarify
• Jun 2nd 2010, 01:47 AM
pickslides
Have you seen partial fractions before?
• Jun 2nd 2010, 03:19 AM
HallsofIvy
No need for partial fractions (writing this as "partial fractions" would require $\frac{A}{x+ 2}+ \frac{B}{(x+ 2)^2}$ 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 $\int \frac{du}{u^2}= \int u^{-2}du$
• Jun 2nd 2010, 05:13 AM
Punch
does this means i do not use ln rule to integrate but rather use normal power rule to intergrate?
• Jun 2nd 2010, 12:57 PM
harish21
Quote:

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.

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

Follow what HallsofIvy has said in the above post.