# Evaluate $\int{\frac{4x}{(x^2-1)(x-1)}dx}$

So I know I need to use the partial fractions method to solve this integral. However when I split it as:

$$\frac{4x}{(x^2-1)(x-1)} = \frac{Ax + B}{x^2-1} + \frac{C}{x-1}$$

I find that I can't solve for the values of A, B, C. The question actually hints that first I need to 'factor the denominator completely'.

I thought it was already factored, but the solution I have factors it to:

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

Can someone please explain the thought-process behind this transformation, why it is more 'completely factored', and why my initial set-up is incorrect?

Question and solution originally from:

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4b.pdf

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/exams/prexam4bsol.pdf

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Your initial setup is incorrect, as there can never be constants $A,B,C$ for which

$$\frac{4x}{(x^2 -1)(x-1)} = \frac{Ax+B}{x^2 -1} + \frac{C}{x-1}$$

This you can see by multiplying by $x^2-1$. You will see that the right side becomes a polynomial, while the left side does not.

The standard techniques of partial fractions try to get the denominator in the form $$(x-a_1)^{r_1} (x - a_2)^{r_2} ... (x-a_n)^{r_n}$$

with the $\displaystyle a_i$ being distinct: this is crucial.

So in your case, $\displaystyle (x^2-1)(x-1)$ becomes $\displaystyle (x-1)(x+1)(x-1) = (x-1)^2(x+1)$

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Another way of splitting the integrals. Write $4x = (x+1)^{2} - (x-1)^{2}$ then you have

\begin{align*} \int\frac{4x}{(x^{2}-1)\cdot(x-1)} \ dx &= \int \frac{ (x+1)^{2}}{(x^{2}-1) \cdot (x-1)} \ dx - \int \frac{(x-1)^{2}}{(x^{2}-1) \cdot (x-1)} \ dx \\ &=\int \frac{x+1}{(x-1)^{2}} \ dx - \int \frac{1}{x+1} \ dx \\ &=\int \frac{1}{x-1} \ dx + 2 \int\frac{1}{(x-1)^{2}} \ dx - \int\frac{1}{x+1} \ dx \end{align*}

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You need to factor the $x^2-1$ term. That will give you a term $(x-1)^2$ in the denominator, so then you need $\displaystyle\frac{A}{(x+1)}+\frac{B}{(x-1)^{1}}+\frac{C}{(x-1)^2}.$

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$(x^2-1)(x-1)$ is not completely factored because $x^2-1$ can be further factored as $(x-1)(x+1)$.

So we get $(x-1)(x+1)(x-1)$. That's completely factored. Since one of the factors appears twice, it's written as something squared.

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