# 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!

• Your partial fractions expansion is incorrect, you should have $B=-4$. – youler Mar 2 '14 at 20:11
• user124539: In case you don't know about accepting an answer to the questions you ask: you can accept at most one answer per question asked. To accept an answer, simply click on the grey $\large \checkmark$ to the left of the answer you'd like to accept. It turns green when you click on it. And you receive 2 reputation points for each accepted answer. – amWhy Mar 3 '14 at 13:37

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

$$\int \left(-\dfrac{1}{x - 1} - \dfrac{4}{x + 1} + \dfrac{1}{(x + 1)^2} \right)\,dx$$

$$-\ln|x - 1| - 4\ln|x + 1| - \dfrac{1}{x + 1} + \mbox{C}$$