Good Day,

I'm faced with a partial fractions and integration problem I've attached. I got the partial fractions part right.But I just can't get the final answer for the integration part.

Help need. Thanks.

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- Dec 13th 2010, 07:45 AMdd86Partial Fractions/ Integration
Good Day,

I'm faced with a partial fractions and integration problem I've attached. I got the partial fractions part right.But I just can't get the final answer for the integration part.

Help need. Thanks. - Dec 13th 2010, 08:33 AMFernandoRevilla
For example,

$\displaystyle -\dfrac{1}{4}\displaystyle\int \dfrac{dx}{-1-x}=\dfrac{1}{4}\left[\log |x+1|\right]_{-10}^{-8}=\dfrac{1}{4}(\log 7-\log 9)=\dfrac{1}{4}\log \dfrac{7}{9}\quad[1]$

$\displaystyle \dfrac{1}{4}\displaystyle\int \dfrac{dx}{1-x}=\ldots =\dfrac{1}{4}\log \dfrac{11}{9}\quad [2]$

Now, $\displaystyle [1]+[2]$ ...

Fernando Revilla - Dec 13th 2010, 03:53 PMdd86
Thank you.

I was half asleep when I posted this, so I didn't actually state what I had a problem with exactly.

When I expressed it in partial fractions, I expressed it as (1/4) / (x+1) instead of (-1/4) / (-1-x).

However, when evaluating it in the integration part, I first left that part out as a log or ln for a negative number does not exist.

Then, after I saw the answer, I tried by taking the -1 out of (x+1). In the end, I ended up with (1/4) ln (81/77).

Now I realise that I left out one tiny thing that messed up my calculations (Headbang) .... the modulus sign (Angry)