For \(x \in \mathrm{R}-\{-1,0,1\}\), \(\int\displaystyle\frac{1}{x-x^5}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right|+c\): where \(c\) is constant of integration.
Since \(x - x^5 = x(1 - x^4)\), multiplying the numerator and denominator by \(x^3\) gives
\( \displaystyle \int\frac{1}{x-x^5}\,dx = \int\frac{x^3}{x^4(1-x^4)}\,dx \)
By substitution, let \(t = x^4\), so \(dt = 4x^3\,dx\):
\( \displaystyle = \frac{1}{4}\int\frac{dt}{t(1-t)} \)
By partial fractions:
\( \displaystyle \frac{1}{t(1-t)} = \frac{1}{t} + \frac{1}{1-t} \)
So
\( \displaystyle \frac{1}{4}\int\left(\frac{1}{t} + \frac{1}{1-t}\right)dt = \frac{1}{4}\Big(\log_e|t| - \log_e|1-t|\Big) + c \)
\( \displaystyle = \frac{1}{4}\log_e\left|\frac{x^4}{1-x^4}\right| + c \)