1. ## integral using substitution

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...

2. 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...

3. hmm the answer are all in terms of u

4. 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

5. 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 $u= \sqrt {1+x}$
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 $u= \sqrt {1+x}$