Partial fraction integration question help

I'm asked first asked to use partial fractions to express this equation.
I've also shown the values of A,B respectively.
http://i.gyazo.com/cb725d61273e20290b7b3c80c454522b.png


The 2nd part asks me to integrate the equation with limits -1 and 0 and to show my answer as pLn(2) were p is a rational number. So I used my partial fraction and integrated it and got the following.


[3ln(2x+1) + 5ln(2-6x)] with limits -1, 0.


So when I sub in -1, I notice I get a negative natural log which can't be done. Have I gone wrong somewhere ?

You've forgotten your modulus signs!

Don't forget that when you integrate something like 1/x you get ln |x| + c, NOT ln x + c

So how would I go abouts calculating this with my calculator ?

Well |x| just means replace x with -x whenever it is negative. So whenever you have a negative number appearing inside a logarithm, just replace it with the corresponding positive number.

Thanks

Just to be sure, is my integral correct ? and so I just input 1 instead of -1 ?

I'm a bit confused by the partial fraction you posted in your original image. How does 2 + x on one side turn into 2x + 1 on the other? Is it supposed to be 2 + x on both sides or 2x + 1 on both sides?

2+x

Sorry that was a type. It should be 2+x on both sides

Then I don't get your partial fraction!

I get

$\dfrac{5 - 8x}{(2+x)(1-3x)} = \dfrac{3}{2 + x} + \dfrac{1}{1 - 3x}$

Just tried it again and got that

How do I show my answer in terms of pln(2), were p is rational, given the boundries of -1 and 0 ?

The integral I got was [3ln(2+x) + ln(1-3x)]

Check your 2nd term - I think you need a factor outside the ln (what would be the derivative of ln(1-3x)?)

Then you just plug the limits in - I don't see any problems with negative numbers appearing inside the logs

Thanks, from this I get a final answer of 11/3 ln(2) ?

I get the same answer