Here is the question

http://farm5.static.flickr.com/4098/...104950cfb6.jpg

This is my attempt. Is this correct?

http://farm5.static.flickr.com/4117/...bc4f5b58_z.jpg

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- Jul 22nd 2010, 03:03 AMracewithferrariIntegration Problem....answer check???
**Here is the question**

http://farm5.static.flickr.com/4098/...104950cfb6.jpg

**This is my attempt. Is this correct?**

http://farm5.static.flickr.com/4117/...bc4f5b58_z.jpg - Jul 22nd 2010, 03:39 AMGeneral
Correct :) .

- Jul 22nd 2010, 03:46 AMTheCoffeeMachine
Correct.

You could have also used the substitution $\displaystyle u = x+3$:

$\displaystyle \displaystyle \blacktriangleright 3\int\frac{r^2}{r+3}\;{dr} = \int\frac{(u-3)^2}{u}\;{du}$

$\displaystyle \displaystyle = 3\int\frac{u^2-6u+9}{u}\;{du} = 3\int{\frac{9}{u}+u-6\;{du}$

$\displaystyle \displaystyle = 27\ln{u}-\frac{3}{2}u^2-18u+k = 27\ln(r+3)-\frac{3}{2}(r+3)^2-18(r+3)+k$ - Jul 22nd 2010, 03:49 AMGeneral
^^^

Its r not x. - Jul 22nd 2010, 05:20 PMracewithferrari
Thanks. But the correct answer include abs sign between ln.