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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
Dear Ryan, Welcome to math.SE. since you are a new user, we wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Further, it would be better if you could typeset your problem so that it is easy for people to read. Kindly look… for more details. – user17762 Oct 28 '12 at 6:10
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.

To check, move your mouse over the following:

$$\frac{5}{x+1}+\frac{5}{(x+1)^2}+\frac{4}{x-1} = \frac{9x^2+13x-6}{(x-1)(x+1)^2}$$

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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.

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