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\(\int \frac{1}{x(1 + x^{4})}dx\) is equal to

  • A.\(\frac{1}{5}\log_{e}\left|\frac{x}{1 + x^{4}}\right| + C\)
  • B.\(\frac{1}{5}\log_{e}|1 + x^{4}| + C\)
  • C.\(\frac{1}{4}\log_{e}|1 + x^{4}| + C\)
  • D.\(\frac{1}{4}\log_{e}\left|\frac{x}{1 + x^{4}}\right| + C\)
  • E.\(\frac{1}{4}\log_{e}\left|\frac{x^{4}}{1 + x^{4}}\right| + C\)

Answer: E

The integral is \(\frac{1}{4}\log_{e}\left|\frac{x^{4}}{1 + x^{4}}\right| + C\).

Multiply top and bottom by \(x^{3}\) and put \(t = x^{4}\), \(dt = 4x^{3}dx\):

\[ \begin{aligned} \int\frac{x^{3}}{x^{4}(1 + x^{4})}dx &= \frac{1}{4}\int\frac{dt}{t(1 + t)} \\ &= \frac{1}{4}\int\left(\frac{1}{t} - \frac{1}{1 + t}\right)dt \\ &= \frac{1}{4}\log\left|\frac{t}{1 + t}\right| + C \end{aligned} \]

The options with \(\frac{x}{1 + x^{4}}\) put \(x\) where \(x^{4}\) belongs, and the ones with \(\log|1 + x^{4}|\) alone lose the \(\frac{1}{t}\) part.

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