# How do you integrate int (x^3 - 2) / (x^4 - 1) using partial fractions?

Jul 12, 2017

$\frac{1}{4} \ln | \frac{\left({x}^{2} + 1\right) {\left(x + 1\right)}^{3}}{x - 1} | + {\tan}^{-} 1 \left(x\right) + C$

#### Explanation:

First, let's factor ${x}^{4} - 1$.

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

Using partial fraction separation, we can say that:

$\frac{{x}^{3} - 2}{\left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C x + D}{{x}^{2} + 1}$

${x}^{3} - 2 = A \left(x + 1\right) \left({x}^{2} + 1\right) + B \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) \left(x + 1\right) \left(x - 1\right)$

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Both the $A$ and $C x + D$ terms have a $\left(x + 1\right)$ factor, so we can let $x$ equal $- 1$ and solve for $B$.

${\left(- 1\right)}^{3} - 2 = A \left(0\right) \left(2\right) + B \left(- 2\right) \left(2\right) + \left(- C + D\right) \left(0\right) \left(- 2\right)$

$- 3 = - 4 B$

$\frac{3}{4} = B$

Now we can substitute this into the original equation.

${x}^{3} - 2 = A \left(x + 1\right) \left({x}^{2} + 1\right) + \frac{3}{4} \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) \left(x + 1\right) \left(x - 1\right)$

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Substitute $x = 1$ to get rid of $\left(C x + D\right)$.

$1 - 2 = A \left(2\right) \left(2\right) + \frac{3}{4} \left(0\right) \left(2\right) + \left(C + D\right) \left(2\right) \left(0\right)$

$- 1 = 4 A$

$- \frac{1}{4} = A$

Now substitute this into the original equation.

${x}^{3} - 2 = - \frac{1}{4} \left(x + 1\right) \left({x}^{2} + 1\right) + \frac{3}{4} \left(x - 1\right) \left({x}^{2} + 1\right) + \left(C x + D\right) \left(x + 1\right) \left(x - 1\right)$

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Plug in $x = 0$ to get rid of $C$.

$- 2 = - \frac{1}{4} \left(1\right) \left(1\right) + \frac{3}{4} \left(- 1\right) \left(1\right) + \left(0 C + D\right) \left(1\right) \left(- 1\right)$

$- 2 = - \frac{1}{4} - \frac{3}{4} - D$

$- 1 = - D$

$1 = D$

Now substitute this into the original equation:

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

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Now we can pretty much plug in any number to solve for $C$. Let's do $x = 2$ since that won't make any terms zero.

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

$6 = - \frac{15}{4} + \frac{15}{4} + \left(6 C + 3\right)$

$6 = 6 C + 3$

$3 = 6 C$

$\frac{1}{2} = C$

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So we have:

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

All that is left to do is take the integral of this function.

$\int \left[\frac{- \frac{1}{4}}{x - 1} + \frac{\frac{3}{4}}{x + 1} + \frac{\frac{1}{2} x + 1}{{x}^{2} + 1}\right] \mathrm{dx}$

$= \int \frac{- \frac{1}{4}}{x - 1} \mathrm{dx} + \int \frac{\frac{3}{4}}{x + 1} \mathrm{dx} + \int \frac{\frac{1}{2} x + 1}{{x}^{2} + 1} \mathrm{dx}$

We can integrate the first two pretty easily, and then we can split the numerator of the last integral into two.

$= - \frac{1}{4} \ln | x - 1 | + \frac{3}{4} \ln | x + 1 | + \int \frac{\frac{1}{2} x}{{x}^{2} + 1} \mathrm{dx} + \int \frac{1}{{x}^{2} + 1} \mathrm{dx}$

Use the substitution $u = \left({x}^{2} + 1\right) , \text{ } \mathrm{du} = 2 x \mathrm{dx}$:

$= - \frac{1}{4} \ln | x - 1 | + \frac{3}{4} \ln | x + 1 | + \int \frac{\frac{1}{4}}{u} \mathrm{du} + \int \frac{1}{{x}^{2} + 1} \mathrm{dx}$

$= - \frac{1}{4} \ln | x - 1 | + \frac{3}{4} \ln | x + 1 | + \frac{1}{4} \ln \left(u\right) + {\tan}^{-} 1 \left(x\right)$

And finally simplify everything (don't forget $+ C$ !)

$= - \frac{1}{4} \ln | x - 1 | + \frac{1}{4} \ln | {\left(x + 1\right)}^{3} | + \frac{1}{4} \ln | {x}^{2} + 1 | + {\tan}^{-} 1 \left(x\right)$

$= \frac{1}{4} \ln | \frac{\left({x}^{2} + 1\right) {\left(x + 1\right)}^{3}}{x - 1} | + {\tan}^{-} 1 \left(x\right) + C$