# definite integral (logarithmic) completely lost

• Mar 5th 2012, 09:33 AM
NecroWinter
definite integral (logarithmic) completely lost
problem:

Integrate (2x-1)/(x+1) from 0 to 3

the first thing I do is pull out the constant 2. Then the part to integrate can become ln(x-1)-ln(x+1)

the antiderivative of ln(x-1) is:
x(ln(x-1)-1)-ln(1-x)

the antiderivative of ln(x+1) is:
(x+1)ln(x+1)-x

x(ln(x-1)-1)-ln(1-x)-(x+1)ln(x+1)-x

3(ln(3-1)-1)-ln(1-3)-(3+1)ln(3+1)-x

I can already tell this answer is wrong due to the fact that I have negatives in the ln

the final answer should be 6(1-ln2)

any help?
• Mar 5th 2012, 09:59 AM
TheSaviour
Re: definite integral (logarithmic) completely lost
Quote:

Originally Posted by NecroWinter
Then the part to integrate can become ln(x-1)-ln(x+1)

How? I've no idea how you have arrived at this step.

To do your integral note (2x-1)/(x+1) = 2-[3/(x+1)].
• Mar 5th 2012, 11:26 AM
NecroWinter
Re: definite integral (logarithmic) completely lost
Quote:

Originally Posted by TheSaviour
How? I've no idea how you have arrived at this step.

To do your integral note (2x-1)/(x+1) = 2-[3/(x+1)].

its possible I confused the property of logarithms.

Anyway, how do you go from (2x-1)/(x+1) to 2-[3/(x+1)]
• Mar 5th 2012, 02:25 PM
TheSaviour
Re: definite integral (logarithmic) completely lost
[(2x-1)/(x+1)] =[(2x+2-3)/(x+1)] = [2(x+1)-3]/(x+1) = 2(x+1)/(x+1)-[3/(x+1)] = 2-[3/(x+1)].