Partial Fractions

**symstar** · September 14th 2008, 06:15 PM

I need to do this without using complex numbers.

$\int \frac{10}{(x-1)(x^2+9)}dx$

Working with the fraction:
$\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}$
$10=A(x^2+9)+(Bx+C)(x-1)$
$10=(A+B)x^2+(-B+C)x+(9A-C)$
$C=-1; B=-1; A=1$

Back to the integral:
$\int\frac{1}{x-1}dx-\int\frac{x+1}{x^2+9}dx$

Assuming I made it this far correctly, I'm not quite sure what to do with the second integral... obviously the first is $\ln\left|x-1\right|$

**skeeter** · September 14th 2008, 06:20 PM

$\frac{x+1}{x^2 + 9} = \frac{x}{x^2 + 9} + \frac{1}{x^2 + 9}<br />$

integral of the first fraction will be $\ln\sqrt{x^2+9}$

integral of the second will be $\frac{1}{3}\arctan \left(\frac{x}{3}\right)$

**Chris L T521** · September 14th 2008, 06:21 PM

Originally Posted by symstar

I need to do this without using complex numbers.

$\int \frac{10}{(x-1)(x^2+9)}dx$

Working with the fraction:
$\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}$
$10=A(x^2+9)+(Bx+C)(x-1)$
$10=(A+B)x^2+(-B+C)x+(9A-C)$
$C=-1; B=-1; A=1$

Back to the integral:
$\int\frac{1}{x-1}dx-\int\frac{x+1}{x^2+9}dx$

Assuming I made it this far correctly, I'm not quite sure what to do with the second integral... obviously the first is $\ln\left|x-1\right|$

Note that $\int\frac{x+1}{x^2+9}dx=\int\frac{x}{x^2+9}dx+\int \frac{1}{x^2+9}dx$

The first one can be done using substitution, and then the last one has the form of arctangent!

Try to take it from here.

--Chris

**o_O** · September 14th 2008, 06:22 PM

Basically split up your fraction into two:
$\int \frac{x+1}{x^2 + 9} \ dx \ = \ \int \frac{x}{x^2 + 9} \ dx + \int \frac{1}{x^2+9} \ dx$

**11rdc11** · September 14th 2008, 06:22 PM

I didn't check your work but you can use a trig substituion to solve the integral you are having difficulty with.

**symstar** · September 14th 2008, 06:40 PM

Looking back through my text, I realized I breezed over that section a little bit too quickly. Splitting into 2 integrals was what I missed. Anyways, I got the right answer, thanks everyone for your help!

**Jhevon** · September 14th 2008, 06:42 PM

without partial fractions:

$\frac {10}{(x - 1)(x^2 + 9)} = \frac {x^2 + 9 + 1 - x^2}{(x - 1)(x^2 + 9)}$

............................ $= \frac {x^2 + 9}{(x - 1)(x^2 + 9)} + \frac {(1 - x)(x + 1)}{(x - 1)(x^2 + 9)}$

............................ $= \frac 1{x - 1} - \frac {x + 1}{x^2 + 9}$

............................ $= \frac 1{x - 1} - \frac 1{x^2 + 9} - \frac x{x^2 + 9}$

**Krizalid** · September 14th 2008, 08:39 PM

Haha, it seems you started to give short answers, that's good, keep it up!

**Jhevon** · September 14th 2008, 09:12 PM

Originally Posted by Krizalid

Haha, it seems you started to give short answers, that's good, keep it up!

yup! i was like "Krizalid would be so proud of me

"

by the way, congrats on the 2

th post!

Thread: Partial Fractions

LinkBack

Thread Tools

Display