Integrating

**Punch** · Jun 2nd 2010, 12:44 AM

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

**pickslides** · Jun 2nd 2010, 12:47 AM

Have you seen partial fractions before?

**HallsofIvy** · Jun 2nd 2010, 02:19 AM

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

**Punch** · Jun 2nd 2010, 04:13 AM

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

**harish21** · Jun 2nd 2010, 11:57 AM

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.

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