<s>[INST] Partial fractions Im working on some partial fraction calculus 2 homework problems and I am having some issues with one problem. The question reads:

Evaluate the integral $\displaystyle \int\dfrac{2-(5x^2+x)\;dx}{(x-1)(x+1)^2}$

I approached it by splitting it into the form
$$\dfrac{2-(5x^2+x)}{(x-1)(x+1)^2} = \dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{(x+1)^2}.$$
Then I solved for $A,B$ and $C$ to get $A=-1, B=-2$ and $C=1$. Then I plugged those values back into the form I set up and started to solve for
$$\displaystyle \int\dfrac{2-(5x^2+x)\;dx}{(x-1)(x+1)^2} = - \ln|x-1|-2 \ln|x+1|-\dfrac 1{x+1}+C.$$
When I entered that answer to my online homework it said that it was incorrect and I don't understand why. Hopefully someone can point out where I when wrong and possibly explain how to answer this problem correctly. Thank you! 
 [/INST] Your answer is very close except the constant multiplied by $\ln|x + 1|$.  First, look at
$$\dfrac{2 - x - 5x^2}{(x - 1)(x + 1)^2} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$$
Multiply both sides by $(x - 1)(x + 1)^2$ to get
$$2 - x - 5x^2 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)$$
If $x = 1$, then $A = -1$.  If $x = -1$, then $C = 1$.  Finally, $B = -4$ because
$\begin{aligned}
A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) &= Ax^2 + 2Ax + A + Bx^2 - B + Cx - C\\
&= x^2(A + B) + x(2A + C) + (A - B - C)
\end{aligned}$
which implies that if $A + B = -5$, then $B = -4$ (also works for $A - B - C = 2$).  So we get
$$\int \left(-\dfrac{1}{x - 1} - \dfrac{4}{x + 1} + \dfrac{1}{(x + 1)^2} \right)\,dx$$
Thus, the answer is
$$-\ln|x - 1| - 4\ln|x + 1| - \dfrac{1}{x + 1} + \mbox{C}$$
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