# integral via partial fractions

I have to find the integral of this rational function:

$$\frac{x^5}{(x-1)^2(x^2-1)}$$

Since the power in the numerator is bigger than the power in the denominator, I have to divide the first one by the second one. So I divide it and I have $(x+2)$ and the quotient $(4x^3-2x^2-3x-2)$. Now what I do is write:

$$x^5 = x+2 + \frac{4x^3-2x^2-3x-2}{(x-1)^2 (x^2-1)}$$

Now, when I find the integral of the fraction, I have something like $$x^5 = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} + \frac{D}{x+1}.$$

When I try to find $A ,B,C$ and $D$, I can only find $D=1/8$ and not $A, B$ or $C$. How do I find $A$, $B$, and $C$?

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Instead of $x^5 = x+2 + \dfrac{4x^3-2x^2-3x-2}{(x-1)^2 (x^2-1)}$ you need $\dfrac{x^5}{(x-1)^2 (x^2-1)} = x+2 + \dfrac{4x^3-2x^2-3x-2}{(x-1)^2 (x^2-1)}$ – Michael Hardy Jan 21 '13 at 2:26
Javaman, you've edited all the "personality" out of the question! But my thanks to you and the other editor for the formatting. – Gerry Myerson Jan 21 '13 at 2:34

Assuming you have done your algebra correctly, you have arrived at $${4x^3-2x^2-3x-2\over(x-1)^3(x+1)}={A\over x-1}+{B\over(x-1)^2}+{C\over(x-1)^3}+{D\over x+1}$$ and you want to find $A,B,C,D$. Clear fractions to get $$4x^3-2x^2-3x-2=A(x-1)^2(x+1)+B(x-1)(x+1)+C(x+1)+D(x-1)^3$$ Now you have some options. You can multiply out everything on the right side, gather like powers of $x$, and get $4$ equations for your $4$ unknowns by comparing the coefficients of $x^3$, of $x^2$, od $x$, and of $1$ on both sides of the equation.
Alternatively, if you stick in $x=1$ you immediately get the values of $C$; if you stick in $x=-1$ you immediately get the value of $D$. Then you can stick in two more values of $x$, say, $x=0$ and $x=2$, to get two equations relating $A$ and $B$; or you can differentiate and then stick in $x=1$ to get $B$, and look at the coefficient of $x^3$ to get $A$; and so on.