# U-sub integral: Not sure where i went wrong.

1. Jan 21, 2013

### dwdoyle8854

1. The problem statement, all variables and given/known data

Im not sure what i did wrong, every operation i took to evaluate the integral seems valid. I took pictures of the integral evaluated two different ways, producing two different answers. What bugs me is even if i evaluate (as i did) from say 0 to 1 the numbers are different, so they are not equivalent. I dont see any errors in the first one, but the text book lists the latter as a correct answer.

The incorrect way:

Correct:

2. Relevant equations

3. The attempt at a solution
The incorrect way:
<a href="[PLAIN]http://imgur.com/2QfhdVs"><img [Broken] src="http://i.imgur.com/2QfhdVs.jpg?1" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

Correct:
<a href="[PLAIN]http://imgur.com/24E2jat"><img [Broken] src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

Last edited by a moderator: May 6, 2017
2. Jan 21, 2013

### Dick

Saying u/(2+u^2-2u) equals u/2+u^2/u-2u/u is seriously bad algebra. (a+b)/c=a/c+b/c. c/(a+b) IS NOT equal to c/a+c/b. Try some numbers.

3. Jan 21, 2013

### SammyS

Staff Emeritus
Your error is in doing basic algebra with a rational expression.

$\displaystyle \frac{u}{u^2-2u+2}\ne\frac{u}{u^2}-\frac{u}{2u}+\frac{u}{2}$

After all, is $\displaystyle \ \ \frac{1}{1+2+3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ ? \ \$

Last edited by a moderator: May 6, 2017
4. Jan 21, 2013

### dwdoyle8854

oh christ. im dumb.

Apologies.