<s>[INST] 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!
 [/INST] 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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