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

[dwdoyle8854]

Homework Statement

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 don't see any errors in the first one, but the textbook lists the latter as a correct answer.

The incorrect way:

Correct:

Homework Equations

The Attempt at a Solution

The incorrect way:
<a href="[PLAIN]http://imgur.com/2QfhdVs"><img 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 src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

 

[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.

 

[SammyS]

Homework Statement

I'm 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 don't see any errors in the first one, but the textbook lists the latter as a correct answer.

The incorrect way:
[ IMG]http://i.imgur.com/2QfhdVs.jpg?1[/PLAIN]
Correct:
[ IMG]http://i.imgur.com/24E2jat.jpg[/PLAIN]

Homework Equations

The Attempt at a Solution

The incorrect way:
<a href="[PLAIN]http://imgur.com/2QfhdVs"><img 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 src="http://i.imgur.com/24E2jat.jpg" title="Hosted by imgur.com" alt="" /></a>[/PLAIN]

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}\ ? \ \

 

[dwdoyle8854]

oh christ. I am dumb.

Apologies.