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

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- Jun 2nd 2010, 12:44 AMPunchIntegrating
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 - Jun 2nd 2010, 12:47 AMpickslides
Have you seen partial fractions before?

- Jun 2nd 2010, 02:19 AMHallsofIvy
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 $\displaystyle \int \frac{du}{u^2}= \int u^{-2}du$ - Jun 2nd 2010, 04:13 AMPunch
does this means i do not use ln rule to integrate but rather use normal power rule to intergrate?

- Jun 2nd 2010, 11:57 AMharish21