Q.Integrate the following function: $x (\log x)^2$

The key idea is Power Rule Integration combined with substitution. Let $I = \int x (\log x)^2 \, dx$.

Step 1: Set $t = \log x$, so $x = e^t$ and $dx = e^t \, dt$. The integral becomes:

Step 2: Integrate by parts twice. First, let $u = t^2$, $dv = e^{2t} dt$, giving $du = 2t \, dt$, $v = \frac{e^{2t}}{2}$:

Step 3: For $\int t e^{2t} dt$, let $u = t$, $dv = e^{2t} dt$, so $du = dt$, $v = \frac{e^{2t}}{2}$:

Step 4: Substitute back and simplify: …

The core idea here is that we have a product of a polynomial ($x$) and a logarithmic power ($(\log x)^2$). The power rule for integration alone can't handle the logarithm directly — but integration by parts can. Why? Because differentiating a logarithm reduces its power, while integrating a polynomial is straightforward. So we repeatedly peel off the logarithm layer by layer.

Let’s work through it step by step.

1. Set up integration by parts. We use the formula:

Choose $u = (\log x)^2$ (so that differentiating reduces the power) and $dv = x \, dx$ (easy to integrate).

Then:

2. Apply the formula.

Simplify the new integral:

Now we have a simpler integral: $\int x \log x \, dx$. This still has a logarithm, so we apply integration by parts again.

3. Second integration by parts. For $\int x \log x \, dx$, set $u = \log x$ and $dv = x \, dx$. Then:

So:

4. Substitute back. From step 2: …

Method: Integration by parts with a squared logarithm

For $x^{n}(\log x)^{2}$, by parts with the log-power as $u$ reduces $(\log x)^{2}$ to $\log x$, then a second pass finishes it.

Steps

Step 1: Pick $u=(\log x)^{2}$, $dv=x\,dx$ (ILATE: log before algebraic).

Step 2: First pass.

The exponent on $\log x$ has dropped from $2$ to $1$. …

Common Mistakes

Mistake 1: Differentiating $(\log x)^{2}$ as $\dfrac{2}{x}$ (forgetting the chain rule).

Why it's wrong: $\dfrac{d}{dx}(\log x)^{2}=2\log x\cdot\dfrac1x=\dfrac{2\log x}{x}$. Correct approach: keep the inner $\log x$.

Mistake 2: Not simplifying $\dfrac{x^{2}}{2}\cdot\dfrac{2\log x}{x}$ before integrating. …