Q.Find $\int \left[\log(\log x) + \dfrac{1}{(\log x)^2}\right] dx$

We integrate by parts, setting $I = \int \log(\log x) \, dx$ and handling the second term separately.

First, for $I$, let $t = \log x \Rightarrow x = e^t, dx = e^t dt$. Then

$I = \int \log t \cdot e^t dt$. Integrate by parts: let $u = \log t, dv = e^t dt$, so $du = \frac{1}{t} dt, v = e^t$.

Thus $I = e^t \log t - \int \frac{e^t}{t} dt = x \log(\log x) - \int \frac{e^t}{t} dt$.

Now the original integral is

$I + \int \frac{1}{(\log x)^2} dx = x \log(\log x) - \int \frac{e^t}{t} dt + \int \frac{1}{t^2} \cdot e^t dt$. …

The problem asks us to integrate a sum of two seemingly unrelated functions: $\log(\log x)$ and $\frac{1}{(\log x)^2}$. At first glance, neither looks like a standard integral. But the trick lies in noticing that the derivative of something like $x \log(\log x)$ will involve both a $\log(\log x)$ term and a $\frac{1}{\log x}$ term — not quite what we have, but close. The presence of $\frac{1}{(\log x)^2}$ suggests that a second integration by parts might clean things up.

Let’s work through it step by step.

1. Split the integral Write the given integral as the sum of two separate integrals:

2. Focus on the first part: $I_1 = \int \log(\log x) \, dx$ This is a classic candidate for integration by parts. Set:
   • $u = \log(\log x)$ (so $du = \frac{1}{x \log x} \, dx$)
   • $dv = dx$ (so $v = x$) Then:

3. Now we have a new integral: $J = \int \frac{1}{\log x} \, dx$ This is not a standard elementary integral, but we’ll handle it when it reappears. For now, note:

4. Now tackle the second part: $I_2 = \int \frac{1}{(\log x)^2} \, dx$ Again, use integration by parts. Let:
   • $u = \frac{1}{(\log x)^2}$ (so $du = -\frac{2}{x (\log x)^3} \, dx$)
   • $dv = dx$ (so $v = x$) Then:

This seems to be making things worse — we get a higher power in the denominator. But wait, there’s a better way.

5. A smarter approach: combine the integrals Let’s go back to the original sum. Consider the derivative of $x \log(\log x)$:

So:

(which we already had).

Now consider the derivative of $\frac{x}{\log x}$: …

Method: Integration by parts that engineers a cancellation

Use this for integrals of a sum where one piece integrates by parts into a non-elementary integral that a second piece exactly cancels — a common design in $\log(\log x)$-type problems.

Steps

Step 1: Split the sum and integrate the harder term by parts.

For $\int\log(\log x)\,dx$ take $u=\log(\log x)$, $dv=dx$, so $v=x$ and $du=\dfrac{1}{x\log x}\,dx$:

The leftover $\int\dfrac{1}{\log x}\,dx$ is non-elementary — do not panic, keep it symbolic.

Step 2: Express the second term as the derivative of a product.

Notice $\dfrac{d}{dx}\!\left(\dfrac{x}{\log x}\right)=\dfrac{1}{\log x}-\dfrac{1}{(\log x)^2}$. Rearranged, this shows

Step 3: Add the two parts so the awkward integral cancels. …

Common Mistakes

Mistake 1: Writing $\int\dfrac{1}{\log x}\,dx=\log(\log x)$.

Why it's wrong: $\dfrac{d}{dx}\log(\log x)=\dfrac{1}{x\log x}$, not $\dfrac{1}{\log x}$; in fact $\int\dfrac{1}{\log x}\,dx$ is non-elementary. Correct approach: keep it symbolic and let it cancel — never evaluate it to a wrong closed form.

Mistake 2: Panicking and abandoning the problem at the non-elementary integral.

Why it's wrong: the two terms of the sum are deliberately paired so the awkward $\int\tfrac{1}{\log x}\,dx$ cancels. Correct approach: keep both pieces and add them; the troublesome integral drops out. …