\( \int\displaystyle\frac{\log x-3}{(\log x)^4}\,dx \) is equal to
(where \( C \) is an arbitrary constant and \( \log_e x=\log x \) )
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{x}{(\log x)^3}+\mathrm{C} \)
Let \( \log x=t \), so \( x=e^t \) and \( dx=e^t\,dt \):
\( \displaystyle \int\frac{t-3}{t^4}e^t\,dt=\int e^t\left(\frac{1}{t^3}-\frac{3}{t^4}\right)dt \)
This has the form \( \int e^t[f(t)+f'(t)]\,dt=e^tf(t)+C \) with \( f(t)=\frac{1}{t^3} \):
\( \displaystyle =\frac{e^t}{t^3}+C \)
\( \displaystyle =\frac{x}{(\log x)^3}+C \)
Shortcut: Differentiating Option (2) gives back the integrand:
\( \displaystyle \frac{d}{dx}\frac{x}{(\log x)^3}=\frac{(\log x)^3-x\cdot\frac{3(\log x)^2}{x}}{(\log x)^6} \)
\( \displaystyle =\frac{\log x-3}{(\log x)^4} \)