Find the area bounded by the arc of the curve y = 1/(x

Find the area bounded by the arc of the curve $y = \frac{1}{x\ln^2 x}$ and the lines $x = e$, $x = \frac{1}{e}$, and $y = 0$.

We will check the function's behavior at $x=1$ before attempting to integrate across it.

Step 1: Understand the region

The curve $y=\frac{1}{x\ln^2 x}$ is defined for $x>0$, $x\neq1$. The requested region runs from $x=\frac1e$ to $x=e$, bounded below by $y=0$.

Step 2: Check the function's behavior

At $x=\frac1e$: $\ln\frac1e=-1$, so $y=e$. At $x=e$: $\ln e=1$, so $y=\frac1e$. As $x\to1$: $\ln x\to0$, so $y\to+\infty$ — there is a vertical asymptote right inside the interval.

Step 3: Split the integral at the singularity

\[A = \int_{1/e}^{1} \frac{1}{x\ln^2 x} \, dx + \int_{1}^{e} \frac{1}{x\ln^2 x} \, dx\]

Step 4: Find the antiderivative

With $u=\ln x$, $du=\frac1x\,dx$:

\[\int \frac{1}{x\ln^2 x} \, dx = \int \frac{1}{u^2} \, du = -\frac{1}{u} + C = -\frac{1}{\ln x} + C\]

Step 5: Examine each piece as an improper integral

\[\int_{1/e}^{1} \frac{1}{x\ln^2 x} \, dx = \lim_{b\to1^-}\left[-\frac{1}{\ln x}\right]_{1/e}^{b}\]

At $x=\frac1e$: $-\frac{1}{-1}=1$. As $x\to1^-$: $\ln x\to0^-$, so $-\frac{1}{\ln x}\to+\infty$. This piece diverges to $+\infty$.

Similarly:

\[\int_{1}^{e} \frac{1}{x\ln^2 x} \, dx = \lim_{a\to1^+}\left[-\frac{1}{\ln x}\right]_{a}^{e}\]

At $x=e$: $-\frac11=-1$. As $x\to1^+$: $\ln x\to0^+$, so $-\frac{1}{\ln x}\to-\infty$, meaning the piece is $-1-(-\infty)=+\infty$: it also diverges.

Conclusion

\[\boxed{\text{The area is infinite} - \text{the integral diverges at } x=1}\]

The function $\frac{1}{x\ln^2 x}$ behaves like $\frac{1}{(x-1)^2}$ near $x=1$ (since $\ln x\approx x-1$ there), which is a non-integrable singularity: the curve shoots to infinity too quickly on both sides of $x=1$ for the enclosed area to be finite.

A bounded variant that avoids the singularity

If the region is instead taken over an interval not containing $x=1$ — for instance $[e, e^2]$ — the integral is proper:

\[A = \int_{e}^{e^2} \frac{1}{x\ln^2 x} \, dx = \left[-\frac{1}{\ln x}\right]_{e}^{e^2} = -\frac{1}{2} - (-1) = \frac{1}{2}\]

\[\boxed{\text{Over } [e, e^2]: A = \frac{1}{2} \text{ square units}}\]

Summary

For the region from $x=\frac1e$ to $x=e$ (which straddles the singularity at $x=1$):

\[\boxed{A = \infty}\]

For a region avoiding the singularity, such as $[e,e^2]$:

\[\boxed{A = \frac12 \text{ square units}}\]

Key insight: Not every curve with a vertical asymptote bounds a finite area near it — the rate of blow-up matters. Here $\frac{1}{x\ln^2x}\sim\frac{1}{(x-1)^2}$ near $x=1$, the same non-integrable order as $\frac1{x^2}$ at the origin.