# How do you integrate int x/(16x^4-1) using partial fractions?

Apr 13, 2017

$\frac{1}{16} \ln | \frac{4 {x}^{2} - 1}{4 {x}^{2} + 1} | + C$

#### Explanation:

First, factorize the denominator to get $\left(4 {x}^{2} + 1\right) \left(2 x + 1\right) \left(2 x - 1\right)$.

We need to find $A$, $B$, $C$, and $D$ such that $\frac{x}{16 {x}^{4} - 1} = \frac{A x + B}{4 {x}^{2} + 1} + \frac{C}{2 x + 1} + \frac{D}{2 x - 1}$ for all $x$. Multiply both sides by $\left(4 {x}^{2} + 1\right) \left(2 x + 1\right) \left(2 x - 1\right)$.

Then, $x = \left(A x + B\right) \left(2 x + 1\right) \left(2 x - 1\right) + C \left(4 {x}^{2} + 1\right) \left(2 x - 1\right) + D \left(4 {x}^{2} + 1\right) \left(2 x + 1\right)$.

Set $x = \frac{1}{2}$ to get $\frac{1}{2} = D \cdot 2 \cdot 2$, or $D = \frac{1}{8}$.

Set $x = - \frac{1}{2}$ to get $- \frac{1}{2} = C \cdot 2 \cdot - 2$, or $C = \frac{1}{8}$.

Set $x = \frac{i}{2}$ to get $\frac{i}{2} = \left(\frac{A i}{2} + B\right) \left(i + 1\right) \left(i - 1\right) = \left(A \left(\frac{i}{2}\right) + B\right) \cdot - 2$, or $i = - 2 A i - 4 B$. Thus, $A = - \frac{1}{2}$ and $B = 0$.

From the above, it can be seen that $\frac{x}{16 {x}^{4} - 1} = \frac{- \frac{1}{2} x}{4 {x}^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}$.

The problem the becomes $\int \setminus \left(\frac{- \frac{1}{2} x}{4 {x}^{2} + 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{\frac{1}{8}}{2 x - 1}\right) \setminus \mathrm{dx}$, or $- \frac{1}{2} \int \setminus \frac{x}{4 {x}^{2} + 1} \setminus \mathrm{dx} + \frac{1}{8} \int \setminus \frac{1}{2 x + 1} \setminus \mathrm{dx} + \frac{1}{8} \int \setminus \frac{1}{2 x - 1} \setminus \mathrm{dx}$.

For the first integral, use the substitution $u = 4 {x}^{2} + 1$ and $\mathrm{du} = 8 x \setminus \mathrm{dx}$ to get $- \frac{1}{2} \int \setminus \frac{x}{4 {x}^{2} + 1} \setminus \mathrm{dx} = - \frac{1}{16} \int \setminus \frac{1}{u} \setminus \mathrm{du} = - \frac{1}{16} \ln | u | + C$. Substitute $u = 4 {x}^{2} + 1$ to get $- \frac{1}{16} \ln | 4 {x}^{2} + 1 | + C$.

For the second integral, use the substitution $u = 2 x + 1$ and $\mathrm{du} = 2 \setminus \mathrm{dx}$ to get $\frac{1}{8} \int \setminus \frac{1}{2 x + 1} \setminus \mathrm{dx} = \frac{1}{16} \int \setminus \frac{1}{u} \setminus \mathrm{du} = \frac{1}{16} \ln | u | + C$. Substitute $u = 2 x + 1$ to get $\frac{1}{16} \ln | 2 x + 1 | + C$.

For the third integral, use the substitution $u = 2 x - 1$ and $\mathrm{du} = 2 \setminus \mathrm{dx}$ to get $\frac{1}{8} \int \setminus \frac{1}{2 x - 1} \setminus \mathrm{dx} = \frac{1}{16} \int \setminus \frac{1}{u} \setminus \mathrm{du} = \frac{1}{16} \ln | u | + C$. Substitute $u = 2 x - 1$ to get $\frac{1}{16} \ln | 2 x - 1 | + C$.

Combine these to get the final answer $- \frac{1}{16} \ln | 4 {x}^{2} + 1 | + \frac{1}{16} \ln | 2 x + 1 | + \frac{1}{16} \ln | 2 x - 1 | + C = \frac{1}{16} \ln | \frac{4 {x}^{2} - 1}{4 {x}^{2} + 1} | + C$.