Answer: stationary point \(\left(e^{1/3},\dfrac{1}{3e}\right)\); \(\displaystyle\int \dfrac{\ln x}{x^4}\,dx=-\dfrac{1}{9x^3}-\dfrac{\ln x}{3x^3}+C\).
Write
\(\frac{\ln x}{x^3}=x^{-3}\ln x.\)
Differentiate using the product rule:
\(\frac{d}{dx}(x^{-3}\ln x) =x^{-3}\cdot\frac1x+\ln x\cdot(-3x^{-4}).\)
So
\(\frac{d}{dx}\left(\frac{\ln x}{x^3}\right) =x^{-4}-3x^{-4}\ln x =\frac{1-3\ln x}{x^4}.\)
For a stationary point, set the derivative equal to zero:
\(\frac{1-3\ln x}{x^4}=0.\)
Since \(x^4\ne0\),
\(1-3\ln x=0.\)
Therefore
\(\ln x=\frac13,\qquad x=e^{1/3}.\)
The corresponding \(y\)-value is
\(y=\frac{\ln(e^{1/3})}{(e^{1/3})^3} =\frac{1/3}{e} =\frac{1}{3e}.\)
So the stationary point is
\(\left(e^{1/3},\frac{1}{3e}\right).\)
From part (i),
\(\frac{1-3\ln x}{x^4} = \frac1{x^4}-\frac{3\ln x}{x^4}.\)
Hence
\(\frac{\ln x}{x^4} = \frac13\left(\frac1{x^4}-\frac{d}{dx}\left(\frac{\ln x}{x^3}\right)\right).\)
Integrating,
\(\int\frac{\ln x}{x^4}\,dx = \frac13\int x^{-4}\,dx -\frac13\cdot\frac{\ln x}{x^3}.\)
Since \(\int x^{-4}\,dx=-\dfrac{1}{3x^3}\),
\(\int\frac{\ln x}{x^4}\,dx = -\frac{1}{9x^3}-\frac{\ln x}{3x^3}+C.\)
