If ∫dxx4−x2=1x+log⁡|f(x)|+C then f(x) is given by

1x4−x2=1x2(x−1)(x+1)=121x21x−1−1x+1=121x−1−1x−1x2−1x2−1x−1x+1=121x−1−2x2−1x+1∫dxx4−x2=12log⁡|x−1|−log⁡|x+1|+2x+C=1x+log⁡x−1x+11/2+C.

If ∫dxx4−x2=1x+log⁡|f(x)|+C then f(x) is given by

1x4−x2=1x2(x−1)(x+1)=121x21x−1−1x+1=121x−1−1x−1x2−1x2−1x−1x+1=121x−1−2x2−1x+1∫dxx4−x2=12log⁡|x−1|−log⁡|x+1|+2x+C=1x+log⁡x−1x+11/2+C.

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If $\int \frac{d x}{x^{4} - x^{2}} = \frac{1}{x} + \log | f (x) | + C$ then $f (x)$ is given by

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$\frac{x + 1}{x - 1}$

$\frac{x - 1}{x + 1}$

${(\frac{x - 1}{x + 1})}^{1 / 2}$

${(\frac{x - 1}{x + 1})}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{x^{4} - x^{2}} = \frac{1}{x^{2} (x - 1) (x + 1)} = \frac{1}{2} \frac{1}{x^{2}} [\frac{1}{x - 1} - \frac{1}{x + 1}] \\ = \frac{1}{2} [\frac{1}{x - 1} - \frac{1}{x} - \frac{1}{x^{2}} - [\frac{1}{x^{2}} - (\frac{1}{x} - \frac{1}{x + 1})]] \\ = \frac{1}{2} [\frac{1}{x - 1} - \frac{2}{x^{2}} - \frac{1}{x + 1}] \\ \int \frac{d x}{x^{4} - x^{2}} = \frac{1}{2} [\log | x - 1 | - \log | x + 1 | + \frac{2}{x}] + C \\ = \frac{1}{x} + \log {|\frac{x - 1}{x + 1}|}^{1 / 2} + C . \end{array}$