# Simple integration question

• Jul 29th 2011, 05:19 AM
remember
Simple integration question
I'm looking to integrate:

1/(2x+5)

Wolframalpha.com displays the answer as:

1/2 log(2x+5)

My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...

Could someone take me step by step through this?
• Jul 29th 2011, 05:25 AM
Sudharaka
Re: Simple integration question
Quote:

Originally Posted by remember
I'm looking to integrate:

1/(2x+5)

Wolframalpha.com displays the answer as:

1/2 log(2x+5)

My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...

Could someone take me step by step through this?

Hi remember,

$\int\frac{1}{2x+5}dx$

$\mbox{Substitute, }u=2x+5\Rightarrow dx=\frac{du}{2}$

$\int\frac{1}{2u}du=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln|2 x+5|+C$

Therefore, $\int\frac{1}{2x+5}dx=\frac{1}{2}\ln|2x+5|+C$
• Jul 29th 2011, 05:41 AM
HallsofIvy
Re: Simple integration question
Quote:

Originally Posted by remember
I'm looking to integrate:

1/(2x+5)

Wolframalpha.com displays the answer as:

1/2 log(2x+5)

My problem is with the (1/2) in the answer. Wolfram claims that you substitute u=2x+5 and du=2dx, but then goes on to take the 1/2 out before substituting then integrating...

Yes, of course. If c is any constant then $\int cf(x)dx= c\int f(x)dx$

Quote:

Could someone take me step by step through this?
• Jul 29th 2011, 05:42 AM
remember
Re: Simple integration question
Thanks a lot! I really need to brush up on the basics!