Thread: Partial Fractions

I need to do this without using complex numbers.

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

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

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

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

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

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

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

Originally Posted by symstar

I need to do this without using complex numbers.

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

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

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

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

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

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

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

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

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!

without partial fractions:

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

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

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

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

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

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 2th post!