Question

Is Wolfram Alpha confused?
No, I am!

\[\int \frac{5}{2x + 2} dx\]

http://www.wolframalpha.com/input/?i=integrate+5%2F%282x+%2B+2%29

Answer 1

It's correct

Answer 2

If you're confused about log, on the bottom it says:
log(x) is the natural logarithm.

Answer 3

WOLFRAM is always right..ALWAYS

Answer 4

No, that's not that part I'm confused about.

\[u = 2x + 2\]\[du = 2dx\]
\[\frac{5}{2} \int \frac{1}{u} du= \frac{5}{2} \ln |u| + C\]
\[= \frac{5}{2} \ln |2x + 2| + C\]
???

Answer 5

those should be equivalent because of the arbitrary C, wolf just pulled out 2 beforehand.

Answer 6

And wolfram is not always right, im sure there have been cases where it was off..

Answer 7

wolf did this:
(5/2)Int(1/(x+1))

Answer 8

Their own integrals site says the other thing. Now I'm really confused!
http://integrals.wolfram.com/index.jsp?expr=5%2F%282x+%2B+2%29&random=false

Answer 9

I dont see the issue personally.
\[\int\limits_{}^{}\frac{ 5 }{ 2x+2 }dx = \int\limits_{}^{}\frac{ 5 }{ 2(x+1) }dx\]
u = x+ 1
du = dx
\[\frac{ 5 }{ 2 }\int\limits_{}^{}\frac{ 1 }{ u }du \]

Answer 10

\[\frac{ 5 }{ 2 }\ln|x+1| + C\]
And remember, C just obliterates constants. I factored out a constant, you didnt, C just makes up for it. Gotta love uber C power.

Answer 11

Ok so it's basically doing \(\ln 2x\) = \(\ln x + \ln 2\) and sticking \(\ln 2\) in with \(C\)?
I get it now, but that's... weird. :D

Answer 12

5/2 ln |2x+2|+c
=5/2 ln |x+1|+ ( 5/2 ln 2 +c)
=5/2 ln |x+1| +c'
same thing