# Integration question that I 'should' find easy

• May 29th 2009, 06:03 AM
hopingforhelp
Integration question that I 'should' find easy
I was recently doing a past paper and I got myself in a muddle when trying to integrate 2/(2x + 1).

Basically I was just wondering why the integral of the above problem is:
ln(2x + 1) as opposed to 2ln(2x + 1).

I would really appreciate a detailed answer. Thanx :)
• May 29th 2009, 06:09 AM
Rapha
$\int 2/(2x+1) dx = 2 * \int 1/(2x+1)dx$

Substitute 2x+1 =: z, then

$2 \int 1/(2z) dz = \int 1/z dz = ln(z) = ln(2x+1)$

Done
• May 29th 2009, 06:31 AM
hopingforhelp
Thanx, so im guessing the markscheme was wrong to say the integral was just ln(2x+1)?
• May 29th 2009, 09:16 AM
HallsofIvy
Quote:

Originally Posted by Rapha
$\int 2/(2x+1) dx = 2 * \int 1/(2x+1)dx$

Substitute 2x+1 =: z, then

$2 \int 1/z dz = 2*ln(z) = 2*ln(2x+1)$

Done

No. If z= 2x+ 1 then $\frac{dz}{dx}= 2$ so dz= 2dx.

$2\int \frac{dx}{2x+1}= \int \frac{2dx}{2x+1}= \int \frac{dz}{z}$
$ln|z|+ C= ln|2x+1|+ C$