Question

Please help me with the following integrating rational functions by partial fractions problem (click to see).

Answer 1

\[\int\limits_{}^{}\frac{ 2x^2-9x-9 }{ x^3-9x }dx\]

Answer 2

\[2x^2-9x-9=A(x^2-9)+Bx(x-3)+Cx(x+3)\]

Answer 3

B=2
C=-1
How do I solve for A?

Answer 4

I'm willing to bet you used the method where you let x = 3 and x = -3 to eliminate and solve for B and C. You can let x = 0 to eliminate B and C to solve for A.

Alternatively, you can use this method:
\[\frac{2x^2-9x+9}{x(x^2-9)}=\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x-3}\\
2x^2-9x+9=A(x^2-9)+B(x)(x-3)+C(x)(x+3)\\
2x^2-9x+9=(A+B+C)x^2+(-3B+3C)x-9A\]
Matching up the coefficients gives the system
\[\begin{cases}A+B+C=2\\-3C-3C=-9\\-9A=9\end{cases}\]
Then solve for A, B, and C.

Answer 5

*The second equation in the system should be
\[-3B-3C=-9\]

Answer 6

oh right, how did i not see that you set x = 0 to find A? haha
yes i solved for b and c substituting x with -3 and 3

Answer 7

\[\int\limits_{}^{} \left[\frac{ A }{ x }+\frac{ B }{ x-9 } \right]dx \implies A(x-9) + B(x) =2x^2-9x-9 \]We set x to be 9 so we can cancel out A and solve for B.\[A(0)+B(9)=2(9)^2-9(9)-9 \rightarrow B(9) = 72 \rightarrow B=8\]Now we do the same but we this time cancel out B by setting x to 0 and solve for a.\[A(0-9)+B(0)=2(0)^2-9(0)-9 \rightarrow A(-9)=-9 \rightarrow A=1\]We re-write the integral with the partial fractions using A = 1 and B = 8 which looks like this:\[\int\limits_{}^{}\left[ \frac{ 1 }{ x }+\frac{ 8 }{ x-9 } \right]dx=\int\limits_{}^{}\frac{ 1 }{ x }dx+\left( 8*\int\limits_{}^{}\frac{ 1 }{ x-9 }dx \right)=(\ln \left| x \right|+C)+[8(\ln \left| x-9 \right|+C)]\]\[=[(\ln \left| x \right|)+C]+[(\ln \left| x-9 \right|^8)+8C]=\ln \left| x \right|+\ln \left| x-9 \right|^8+9C\]Which simplifies to this:\[=\ln (\left| x \right|*\left| x-9 \right|^8)+9C\]This is the final integral that you get from using partial fractions
@Babyslapmafro

Answer 8

Oh fml. I just noticed the denominator was x^3 - 9x and I thought it was x^2 - 9x. Fml. Whatever lol -.-.

Answer 9

That's why I only had 2 fractions and not 3. Lmfao I only noticed that I had the wrong denominator once I finished all the work. Those 3s are quite misleading; sometimes they look like 2s. All that work for nothing. Well whatever, if your denominator was x^2-9x then that would be the answer lol...