Question

If \(\int \frac{\log (\log x)}{4x(\log x)^{2}}dx=k\left[\frac{1+\log (\log x)}{\log x}\right]+c\), then the value of \(k\) is...

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{2}\)
• C. \(-\frac{1}{2}\)
• D. \(-\frac{1}{4}\)

Hint:

Substitute t equal to log x and then integrate by parts.

Answer 1

The Correct Option is D

Step 1: Differentiate the right side
Let $F=k\cdot\frac{1+\log t}{t}$ with $t=\log x$.
$\frac{dF}{dt}=k\cdot\frac{\frac{1}{t}\cdot t-(1+\log t)}{t^{2}}=-k\frac{\log t}{t^{2}}$.

Step 2: Match with the integrand
In terms of $t$, the integrand is $\frac{\log t}{4t^{2}}$.
So $-k=\frac{1}{4}$.

Step 3: Solve
$k=-\frac{1}{4}$.

Final Answer:
Hence k equals minus one quarter. \[ \boxed{\text{(D) }-\dfrac{1}{4}} \]