$\int \frac{x^{3} - 1}{4 x^{3} - x} d x$ is equal to

Easy

A

$\frac{1}{4} x + \log | x | - \frac{3}{16} \log | 2 x - 1 | - \frac{9}{16} \log | 2 x + 1 | + C$
B

$\frac{1}{4} x - \log | x | - \frac{5}{16} \log | 2 x - 1 | - \frac{7}{16} \log | 2 x + 1 | + C$
C

$\frac{1}{4} x - \log | x | - \frac{7}{16} \log | 2 x - 1 | + \frac{9}{16} \log | 2 x + 1 | + C$
D

$\frac{1}{4} x + \log | x | - \frac{7}{16} \log | 2 x - 1 | - \frac{9}{16} \log | 2 x + 1 | + C$

Solution

$\begin{array}{l} \frac{x^{3} - 1}{4 x^{3} - x} = \frac{1}{4} + \frac{\frac{1}{4} x - 1}{x (2 x - 1) (2 x + 1)} \\ = \frac{1}{4} + \frac{1}{x} - \frac{7}{8} \frac{1}{2 x - 1} - \frac{9}{8} \frac{1}{2 x + 1} \\ \int \frac{x^{3} - 1}{4 x^{3} - x} d x = \frac{1}{4} x + \log | x | - \frac{7}{16} \log | 2 x - 1 | \\ - \frac{9}{16} \log | 2 x + 1 | + C \end{array}$

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