integral using substitution

**Linnus** · January 17th 2008, 04:43 PM

I'm not sure what I did wrong here..
$\int 60x \sqrt {1+x} dx$
$u= \sqrt {1+x}$
$x= u^2-1$
$dx= 2u du$

Then I got

$60 \int (u^3-u) 2u du$

After I integrated I got

$120 (\frac {1} {5} u^5- \frac {1} {3} u^3)$
which isn't one of the answer choices...

**Jhevon** · January 17th 2008, 04:45 PM

Originally Posted by Linnus

I'm not sure what I did wrong here..
$\int 60x \sqrt {1+x} dx$
$u= \sqrt {1+x}$
$x= u^2-1$
$dx= 2u du$

Then I got

$60 \int (u^3-u) 2u du$

After I integrated I got

$120 (\frac {1} {5} u^5- \frac {1} {3} u^3)$
which isn't one of the answer choices...

maybe because your answer is in terms of u when it should be in terms of x...

**Linnus** · January 17th 2008, 04:46 PM

hmm the answer are all in terms of u

**Jhevon** · January 17th 2008, 04:48 PM

Originally Posted by Linnus

hmm the answer are all in terms of u

it's possible that they did a different substitution from you did (u = x + 1 could work, for instance), or you need to simplify your answer to get one of the others, or your answer and the correct choice differ by a constant

what are the answer choices?

**Linnus** · January 17th 2008, 04:52 PM

The closest answer choice is $25u^5-40U^3+C$ ...i wonder if this is a misprint or not. The question told me to used $u= \sqrt {1+x}$

**Jhevon** · January 17th 2008, 04:53 PM

Originally Posted by Linnus

The closest answer choice is $25u^5-40U^3+C$ ...i wonder if this is a misprint or not. The question told me to used $u= \sqrt {1+x}$

yes, it's a misprint, it should be 24u^5

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