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 $\displaystyle \ln\left|x-1\right|$