# Integral - Partial Fraction

• Feb 19th 2008, 01:40 PM
killasnake
Integral - Partial Fraction
Hi I'm trying to solve this Partial Fraction Question: I have found A,B, and C. I just cannot get the correct answer.

$\displaystyle \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx$

$\displaystyle A = -\frac{75}{25}$
$\displaystyle B = 0$
$\displaystyle C = 1$

$\displaystyle \int -\frac{\frac{75}{25}}{x-4} + \int \frac{1}{x^2+9}$

$\displaystyle -3ln(x-4)+\frac{1}{2}ln(x^2+9)+C$
• Feb 19th 2008, 01:43 PM
Jhevon
Quote:

Originally Posted by killasnake
Hi I'm trying to solve this Partial Fraction Question: I have found A,B, and C. I just cannot get the correct answer.

$\displaystyle \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx$

$\displaystyle A = -\frac{75}{25}$
$\displaystyle B = 0$
$\displaystyle C = 1$

$\displaystyle \int -\frac{\frac{75}{25}}{x-4} + \int \frac{1}{x^2+9}$

$\displaystyle -3ln(x-4)+\frac{1}{2}ln(x^2+9)+C$

what were your partial fractions. you do realize that $\displaystyle \frac {-3x^2 + x - 31}{(x - 4)(x^2 + 9)} = \frac A{x - 4} + \frac {Bx + C}{x^2 + 9}$ ?
• Feb 19th 2008, 01:52 PM
killasnake
My Partial Fractions is $\displaystyle \frac A{x - 4} + \frac {Bx + C}{x^2 + 9}$

$\displaystyle -3x^2+1x-31 = A(x^2+9)+(Bx+C)(x-4)$

Then I plugged in the variables and solved for A,B and C

$\displaystyle A = -\frac{75}{25}$

$\displaystyle B = 0$

$\displaystyle C = 1$

A,B and C are correct but I cannot get the final correct answer after getting A,B and C
• Feb 19th 2008, 02:05 PM
Jhevon
Quote:

Originally Posted by killasnake
My Partial Fractions is $\displaystyle \frac A{x - 4} + \frac {Bx + C}{x^2 + 9}$

$\displaystyle -3x^2+1x-31 = A(x^2+9)+(Bx+C)(x-4)$

Then I plugged in the variables and solved for A,B and C

$\displaystyle A = -\frac{75}{25}$

$\displaystyle B = 0$

$\displaystyle C = 1$

A,B and C are correct but I cannot get the final correct answer after getting A,B and C

that is correct. (why not write A as -3 though?)

so now you want to find: $\displaystyle \int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} \right)~dx$

the first is an arctan integral $\displaystyle \left( \mbox{we can write it as } \frac 19 \int \frac 1{\left( \frac x3 \right)^2 + 1}~dx \right)$, the second is a natural log integral. which are you having trouble with?
• Feb 19th 2008, 02:47 PM
killasnake
Well, I solved both intergrals and I believe they are correct. I am just getting frustrated that I am getting the wrong answer. (Headbang)

$\displaystyle \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx$

Integral to Solve:
$\displaystyle \int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} \right)~dx$

Solved Integral:
$\displaystyle \frac{1}{3} arctan(\frac{1}{3}x) - 3 ln(x-4)$
• Feb 19th 2008, 02:52 PM
Jhevon
Quote:

Originally Posted by killasnake
Well, I solved both intergrals and I believe they are correct. I am just getting frustrated that I am getting the wrong answer. (Headbang)

$\displaystyle \int \frac{-3x^2+1x-31}{(x-4)(x^2+9)}dx$

Integral to Solve:
$\displaystyle \int \left( \frac 1{x^2 + 9} - \frac 3{x - 4} \right)~dx$

Solved Integral:
$\displaystyle \frac{1}{3} arctan(\frac{1}{3}x) - 3 ln(x-4)$

well, that's right.

all's well that ends well
• Feb 19th 2008, 03:03 PM
killasnake
Well at least i know i got the right answer. Thank you so much Jhevon!
• Feb 19th 2008, 04:23 PM
Krizalid
(No partial fractions foundation.)

$\displaystyle x - 3x^2 - 31 = x - 4 - 3\left( {x^2 + 9} \right).$