Question

Evaluate $\displaystyle \int \left[\log(\log x) + \frac{1}{(\log x)^{2}}\right] dx$.

Hint:

When you see $\log(\log x)$ or $\dfrac{1}{(\log x)^{2}}$, always try substitution $t = \log x$.

Answer 1

Step 1: Separate the integral. \[ I = \int \log(\log x) \, dx + \int \frac{1}{(\log x)^{2}} \, dx \]

Step 2: Substitution. Let $t = \log x \;\Rightarrow\; dx = e^{t} dt = x dt$. \[ I = \int \log t \cdot e^{t} dt + \int \frac{e^{t}}{t^{2}} dt \]

Step 3: First integral by parts. \[ \int e^{t} \log t \, dt = e^{t} \log t - \int \frac{e^{t}}{t} dt \]

Step 4: Combine both integrals. \[ I = e^{t} \log t - \int \frac{e^{t}}{t} dt + \int \frac{e^{t}}{t^{2}} dt \] Notice that: \[ \frac{d}{dt}\left(-\frac{e^{t}}{t}\right) = -\frac{e^{t}}{t} + \frac{e^{t}}{t^{2}} \] So, \[ - \int \frac{e^{t}}{t} dt + \int \frac{e^{t}}{t^{2}} dt = -\frac{e^{t}}{t} \]

Step 5: Final result. \[ I = e^{t} \log t - \frac{e^{t}}{t} + C \] Back substitute $t = \log x$: \[ I = x \log(\log x) - \frac{x}{\log x} + C \]

Final Answer: \[ \boxed{\; x \log(\log x) - \dfrac{x}{\log x} + C \;} \]