# Partial Fractions - Calculus

Evaluate the integral: $$\int \dfrac{9x^2+13x-6}{(x-1)(x+1)^2} dx$$

For some reason I cannot get the right answer. I split up the equation into three partial fractions but I cannot seem to find A, B, or C from the three subsequent equations. Thanks!

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Can you show us your working? –  wj32 Oct 28 '12 at 6:06
I get (A+B)x^2+(2Ax+Cx)+(A-B-C)=9x^2+13x-6 –  Ryan Oct 28 '12 at 6:09
And then I set each section equal to its respective answer, but cannot figure out how to solve for A, B, or C. –  Ryan Oct 28 '12 at 6:10

From $(A+B) x^2 + (2A+C)x +(A-B-C) = 9x^2+13 x -6$, you need to solve for $A,B,C$. Since the coefficients of the polynomials in $x$ must match, this gives three equations $A+B=9$, $2A+C = 13$, and $A-B-C = -6$. If you solve these you will get your answer.
$$\frac{5}{x+1}+\frac{5}{(x+1)^2}+\frac{4}{x-1} = \frac{9x^2+13x-6}{(x-1)(x+1)^2}$$
The motivation is to write $\dfrac{9x^2 + 13x - 6}{(x-1)(x+1)^2}$ as $$\dfrac{A}{(x+1)^2} + \dfrac{B}{x+1} + \dfrac{C}{x-1}$$ This gives us \begin{align} 9x^2 + 13x - 6 & = A(x-1) + B(x^2-1) + C(x+1)^2\\ & = (B+C)x^2 + (2C+A)x + (C-A-B) \end{align} This gives us $$B+C = 9\\ 2C+A = 13\\ C - A - B = -6$$ Adding all the three equations, give us $4C = 16 \implies C = 4$. Hence, $A=B=5$. Hence, we get that $$\dfrac{9x^2 + 13x - 6}{(x-1)(x+1)^2} = \dfrac5{(x+1)^2} + \dfrac5{x+1} + \dfrac4{x-1}$$ Now you should be able to integrate and finish it off.