1. ## integration help,

 . What is the exact value of ? Give your answer as a fraction or whole number

I know i have to split into partial fractions, but not sure how to go from there

2. ## Re: integration help,

Have you determined the partial fraction decomposition? Once you do, then the integration is straightforward. Are you familiar with the cover-up method?

3. ## Re: integration help,

I have I got $\int^1_0 \frac{-1}{x+1} + \frac{1}{x+2}$

I than get, $\int^1_0 -ln(x+1) + ln(x+2)$

and I dont know how to go from here, i dont even know if am right

4. ## Re: integration help,

That's the correct partial fraction decomposition. In the second line, instead of an integral sign, you want -ln(x+1)+ln(x+2) evaluated from 0 to 1, so it's -ln(1+1)+ln(1+2)+ln(0+1)-ln(0+2). Now you can combine the logarithms to get k.

- Hollywood

5. ## Re: integration help,

Originally Posted by Tweety
I have I got $\int^1_0 \frac{-1}{x+1} + \frac{1}{x+2}$

I than get, $\int^1_0 -ln(x+1) + ln(x+2)$

and I dont know how to go from here, i dont even know if am right
Let me just touch it up a bit:

$\int^1_0 \frac{-1}{(x+1)(x+2)} \ dx$

$= \int^1_0 \left( \frac{-1}{x+1} + \frac{1}{x+2} \right) dx$

$= \int^1_0 \frac{-1}{x+1} \ dx + \int^1_0 \frac{1}{x+2} \ dx$

$= \left\ -ln(x+1)\right]_0^1 + \left\ ln(x+2)\right]_0^1$

And from there:

Spoiler:

$= \left\{ [-ln((1)+1)] - [-ln((0)+1)] \right\} + \left\{ [ln((1)+2)] - [ln((0)+2)] \right\}$

$= \left\{ [-ln(2)] - [-ln(1)] \right\} + \left\{ [ln(3)] - [ln(2)] \right\}$

$= -ln(2) + ln(1) + ln(3) - ln(2)$

$= ln(1) + ln(3)$

$= (0) + ln(3)$

$= ln(3)$

6. ## Re: integration help,

Johnsomeone - could you please check the third and fourth lines of your spoiler?

Thanks,
Hollywood

7. ## Re: integration help,

My spoiler is spoiled!

From $-ln(2) + ln(1) + ln(3) - ln(2)$ to $ln(1) + ln(3)$? Not my best bit of calculation.

It should be

$-ln(2) + ln(1) + ln(3) - ln(2) = ln(3)-2ln(2) = ln(3) - ln(2^2) = ln(3/4)$.

Thanks for catching my error.