If the integral \(I =\int\displaystyle\left\{\log_e(\log_e x)^2+\frac{\alpha}{\log_e x}\right\}dx=x\log_e(\log_e x)^2+C\), where C is constant of integration. Then the value of \(\alpha\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
Here \(\log_e(\log_e x)^2\) means \(\log_e\left[(\log_e x)^2\right]\), the logarithm of the square. Since differentiation reverses integration, the derivative of the right-hand side must equal the integrand.
By the chain rule:
\( \displaystyle \frac{d}{dx}\log_e(\log_e x)^2 = \frac{1}{(\log_e x)^2}\cdot 2\log_e x\cdot\frac{1}{x} \)
\( \displaystyle \frac{d}{dx}\log_e(\log_e x)^2 = \frac{2}{x\log_e x} \)
By the product rule:
\( \displaystyle \frac{d}{dx}\left[x\log_e(\log_e x)^2\right] = \log_e(\log_e x)^2 + x\cdot\frac{2}{x\log_e x} \)
\( \displaystyle \frac{d}{dx}\left[x\log_e(\log_e x)^2\right] = \log_e(\log_e x)^2 + \frac{2}{\log_e x} \)
Comparing with the integrand \(\log_e(\log_e x)^2 + \frac{\alpha}{\log_e x}\):
\(\alpha = 2\)