Question

Help with partial fractions integrals? I keep getting weird coefficients...(problem below)

Answer 1

The site keeps kicking me off... but anyway my original problem was \[\int\limits_{}^{}\frac{ 10 }{ (x-1)(x^2+9) }dx\] and I set up the partial fractions like this \[10=A(x^2+9)+B(x-1)\] but the coefficients don't add up, I get A and B both equalling 0 yet somehow 9A-B equalling 10. How do I do this?

Answer 2

I think you set it up like this
\[ \frac{A}{x-1} + \frac{Bx + C}{x^2 +9} \]
putting these over the common denominator (x-1)(x^2+9) we get
\[ A(x^2+9) + (Bx+C)(x-1) \]
it looks like you left out the C
expanding, and collecting powers of x, and equating to 10
\[ Ax^2 +9A +Bx^2 -Bx + Cx -C = 10 \\
(A+B) x^2 + (C-B)x + (9A-C) = 10
\]
from which we get
A+B=0, C-B=0 , 9A-C = 10

Answer 3

oh because its a quadratic on the bottom, ok thank you :)