# Math Help - [SOLVED] Definite Integral, &amp; Partial fractions

1. Okay, yea, once I got the division right everything went well, I know what to do now.
So I now have:
$\int_{-1}^{3} \frac {85}{x+7}- \frac {6}{x+3}dx$ Could I split the integral, integrate and do the limits individually?
Thanks!

2. Yes and here is a hint $\int\frac{du}{u} = ln(u)$

3. Okay, so I did that and I get 36.8285 for the answer, I then calculated it out on my calculator and it got came out to $85ln(5)-91ln(3)-36$ which is .8285 only, so it's just without the 36. I guess I did something wrong with A and B? I also know that .8285 is the correct answer. I just checked.

4. Oh wait, what do I need to do with the (x-10) on the outside, that must equal -36 right?

5. $\int_{-1}^{3} \bigg(\frac {85}{x+7}- \frac {6}{x+3}\bigg)dx$

$[85ln(x+7) - 6ln(x+3)]^3_{-1}$

so

$85ln(10) - 6ln(6) - 85ln(6) + 6ln(2) = ?$

6. Originally Posted by 11rdc11
Yes and here is a hint $\int\frac{du}{u} = ln(u)$
No, it's $\ln |u|$ .. it doesn't matter here, but it's not very correct to write it

7. Originally Posted by Moo
No, it's $\ln |u|$ .. it doesn't matter here, but it's not very correct to write it
Lol you too picky but you are right!

8. I don't mean to be Mr. Picky, but:

$\left(x-10 + \frac{79x+213}{x^2+10x+21}\right) \times (x^2+10x+21)$

$= (x-10)(x^2+10x+21) + 79x + 213$

$= x^3 + 10x^2 + 21x - 10x^2 - 100x - 210 + 79x + 213$

$= x^3 + 3$

$\therefore \frac{x^3+3}{x^2+10x+21} = x-10 + \frac{79x+213}{x^2+10x+21}$

No need to change the integral answers though. The OP should be able to solve them on his own by now...right Matt3D?

9. Originally Posted by 11rdc11
$\int_{-1}^{3} \bigg(\frac {85}{x+7}- \frac {6}{x+3}\bigg)dx$

$[85ln(x+7) - 6ln(x+3)]^3_{-1}$

so

$85ln(10) - 6ln(6) - 85ln(6) + 6ln(2) = ?$
Yea, that is what I got which is 36.8285

10. Originally Posted by 11rdc11
Yep the first term after doing long division can't be $x^2$. It would have to be x because x times $x^2$ equals $x^3$. Now if the first term was $x^2$ we would have $x^2$ times $x^2$ which would be $x^4$

Oops stand corrected! I didn't notice matt3d corrected his long division and he is correct it 213 not 210. Well everything Jameson said to do after with the partial fractions is correct. We all made minor mistakes lol.
Lol yea we had a couple of mistakes but it all worked out in the end

11. Originally Posted by Chop Suey
I don't mean to be Mr. Picky, but:

$\left(x-10 + \frac{79x+213}{x^2+10x+21}\right) \times (x^2+10x+21)$

$= (x-10)(x^2+10x+21) + 79x + 213$

$= x^3 + 10x^2 + 21x - 10x^2 - 100x - 210 + 79x + 213$

$= x^3 + 3$

$\therefore \frac{x^3+3}{x^2+10x+21} = x-10 + \frac{79x+213}{x^2+10x+21}$

No need to change the integral answers though. The OP should be able to solve them on his own by now...right Matt3D?
Yea, I know what you mean, it's all good though, I just thought that I should point it out... I'm still stuck on the answer though.

12. Originally Posted by matt3D
Yea, that is what I got which is 36.8285

Ok I figured out why it wrong we forgot the other part of the original integral lol

$\int_{-1}^3 \bigg(x - 10 + \frac{85}{x+7} - \frac{6}{x+3}\bigg)dx$

so

$\frac{x^2}{2} -10x + 85ln(x+7) - 6ln(x+3)\bigg|^3_{-1}$

13. Haha, awesome, I thought it was from the (x-10) too. Yep, so its 8/2-(30+10) which is -36.

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