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

Aug 29, 2017

$x + \frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \arctan \frac{x}{2} + C$

#### Explanation:

Before we rewrite ${x}^{4} / \left({x}^{4} - 1\right)$ into partial fractions, first make sure that the degree of the denominator is strictly greater than the degree of the numerator.

So, we can use ${x}^{4} / \left({x}^{4} - 1\right) = \frac{{x}^{4} - 1 + 1}{{x}^{4} - 1} = 1 + \frac{1}{{x}^{4} - 1}$

Now, we rewrite $\frac{1}{{x}^{4} - 1}$ using partial fractions. First, factorize ${x}^{4} - 1 = \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)$.

Thus, we know that $\frac{1}{{x}^{4} - 1}$ can be rewritten in the form $\frac{A x + B}{{x}^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$.

Multiply both sides by ${x}^{4} - 1$ to get $1 = \left(A x + B\right) \left(x + 1\right) \left(x - 1\right) + C \left({x}^{2} + 1\right) \left(x - 1\right) + D \left({x}^{2} + 1\right) \left(x + 1\right)$.

Set $x = 1$ to get $D = \frac{1}{4}$ and $x = - 1$ to get $C = - \frac{1}{4}$

Multiply this out to get $\left(A + C + D\right) {x}^{3} + \left(B - C + D\right) {x}^{2} - \left(A - C - D\right) x - B - C + D = 1$.

Substitute $D = \frac{1}{4}$ and $C = - \frac{1}{4}$ to finally get $A {x}^{3} + \left(B + \frac{1}{2}\right) {x}^{2} - A x - B + \frac{1}{2} = 1$.

Equate coefficients: $\left\{\begin{matrix}A = 0 \\ B + \frac{1}{2} = 0 \\ A = 0 \\ - B + \frac{1}{2} = 1\end{matrix}\right.$.

From the above, we get $\left\{\begin{matrix}A = 0 \\ B = - \frac{1}{2} \\ C = - \frac{1}{4} \\ D = \frac{1}{4}\end{matrix}\right.$.

Thus, we know that $\frac{1}{{x}^{4} - 1} = - \frac{1}{2 {x}^{2} + 2} - \frac{1}{4 x + 4} + \frac{1}{4 x - 4}$.

Returning to the original problem, we can now integrate $\int \setminus 1 - \frac{1}{2 {x}^{2} + 2} - \frac{1}{4 x + 4} + \frac{1}{4 x - 4} \setminus \mathrm{dx}$.

The first, third, and fourth can be easily solved using either the constant rule or u-substitution: $\int \setminus \mathrm{dx} = x$, $- \int \setminus \frac{\mathrm{dx}}{4 x + 4} = - \frac{\ln | x + 1 |}{4}$, and $\int \setminus \frac{\mathrm{dx}}{4 x - 4} = \frac{\ln | x - 1 |}{4}$.

The second one, $- \int \setminus \frac{\mathrm{dx}}{2 {x}^{2} + 2} = - \frac{1}{2} \int \setminus \frac{\mathrm{dx}}{{x}^{2} + 1}$, requires knowing the identity $\int \setminus \frac{\mathrm{dx}}{{x}^{2} + 1} = \arctan \left(x\right)$.

Thus, we can combine everything to get the final answer: $x + \frac{1}{4} \ln | \frac{x - 1}{x + 1} | - \arctan \frac{x}{2} + C$.