# Integrating Fractions with Substitution

1. Jun 3, 2013

### cp255

So the problem is ∫(6x+5)/(2x+1)dx. I know the proper way to solve this is to long divide these two expressions and then solve. However, I tried doing it with substitution.

u = 2x+1
dx = du/2
I then reasoned that 3u + 2 = 6x+5 since 3(2x+1) + 2 = 6x+3+2 = 6x+5 so I substituted it on top.
(1/2)∫(3u+2)/u du
Solving this integral gives me (3/2)u + ln|u| + C which equals (6x+3)/2 + ln|2x+1| + C.

However, Wolfram Alpha says the answer is 3x + ln|2x+1| + C. I don't understand what I did wrong or why I can't do what I did.

2. Jun 3, 2013

### scurty

It's the same answer. The 3/2 term gets absorbed into the + C term because 3/2 is just a constant. You did the problem correctly.

3. Jun 3, 2013

### cp255

Well that makes sence. It is sometimes hard to check my calculus work especially with computers since there are so many different forms an answer can take.

4. Jun 3, 2013

### scurty

A useful trick I do for integration problems is plug the answer in and take the derivative. If you get the original problem back, you have the correct answer. Wolfram doesn't always compute it the same way as you and might simplify it differently. This is especially apparent if you do a trig substitution and have a theta hanging around in your answer. Using the triangle you made for the trig substitution, you can have 6 different answers for the value of theta that are the same (arcsin, arccos, etc).

You can also differentiate by hand if the answer isn't too complicated to check yourself.