1. ## Simple integration question

I'm looking to integrate:

1/(2x+5)

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?

2. ## Re: Simple integration question

Originally Posted by remember
I'm looking to integrate:

1/(2x+5)

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,

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

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

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

Therefore, $\displaystyle \int\frac{1}{2x+5}dx=\frac{1}{2}\ln|2x+5|+C$

3. ## Re: Simple integration question

Originally Posted by remember
I'm looking to integrate:

1/(2x+5)

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