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

Sep 19, 2016

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

#### Explanation:

We see that:

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

Split it up into its partial fractions:

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

Multiplying through, we see that:

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

Continue:

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

${x}^{3} - 2 = A {x}^{3} - A x + B {x}^{2} - B + C {x}^{3} - C {x}^{2} + C x - C + D {x}^{3} + D {x}^{2} + D x + D$

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

Equating the coefficients on either side of the equation:

$\left\{\begin{matrix}1 = A + C + D \\ 0 = B - C + D \\ 0 = - A + C + D \\ - 2 = - B - C + D\end{matrix}\right.$

Adding the first and third equations, we see that $1 = 2 C + 2 D$. Adding the second and fourth equations, we see that $- 2 = - 2 C + 2 D$. Adding these two equations, this yields $- 1 = 4 D$, so $D = - \frac{1}{4}$.

Substituting $D = - \frac{1}{4}$ into $1 = 2 C + 2 D$, we see that $C = \frac{3}{4}$.

Substituting these values into $1 = A + C + D$, this yields that $A = \frac{1}{2}$.

Furthermore, since $0 = B - C + D$, we see that $B = 1$.

Thus:

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

So:

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

Split up the first integral. The last two integrals can be integrated simply:

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

Rearranging the first term to set up for a natural logarithm substitution, since $2 x$ is the derivative of ${x}^{2} + 1$:

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

The second integral is a common integral as well:

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