\( \int\displaystyle\frac{1}{x(x^6+1)}\,dx= \) (consider \( \log_e x=\log x \) )
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{6}\log\left|\displaystyle\frac{x^6}{x^6+1}\right|+C \) , where \( C \) is an arbitrary constant
Multiplying the numerator and the denominator by \( x^5 \) brings in the derivative of \( x^6 \):
\( \displaystyle \int\frac{dx}{x(x^6+1)}=\int\frac{x^5}{x^6(x^6+1)}\,dx \)
Put \( x^6=t \), so \( 6x^5\,dx=dt \):
\( \displaystyle \int\frac{x^5}{x^6(x^6+1)}\,dx=\frac{1}{6}\int\frac{dt}{t(t+1)} \)
\( \displaystyle =\frac{1}{6}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt \)
\( \displaystyle =\frac{1}{6}\log\left|\frac{t}{t+1}\right|+C \)
\( \displaystyle =\frac{1}{6}\log\left|\frac{x^6}{x^6+1}\right|+C \)
Shortcut: Differentiating Option (1):
\( \displaystyle \frac{1}{6}\left(\frac{6x^5}{x^6}-\frac{6x^5}{x^6+1}\right) \)
\( \displaystyle =\frac{1}{x}-\frac{x^5}{x^6+1} \)
\( \displaystyle =\frac{1}{x(x^6+1)} \)
which is the integrand.