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

Concept: Integration by parts (LIATE rule — log comes before algebraic).

We want $\int x \log(2x) \, dx$.

Step 1: Choose $u = \log(2x)$ and $dv = x \, dx$.

Then $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.

Step 2: Apply integration by parts:

$\int u \, dv = uv - \int v \, du$

$= \frac{x^2}{2} \log(2x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx$

$= \frac{x^2}{2} \log(2x) - \frac{1}{2} \int x \, dx$.

Step 3: Integrate the remaining term: …

Why integration by parts?

When you see a product of two different kinds of functions — here, a polynomial ($x$) and a logarithm ($\log 2x$) — the standard tool is integration by parts. The formula is:

The trick is choosing which part becomes $u$ and which becomes $dv$. For products involving a logarithm, a reliable rule of thumb is: let $u$ be the logarithmic function, because its derivative simplifies to a rational function. The polynomial part then becomes $dv$, which is easy to integrate.

So here, set:

• $u = \log 2x$
• $dv = x \, dx$

Let’s work through it.

Step-by-step solution

1. Identify $u$ and $dv$

We choose:

2. Differentiate $u$ to get $du$

Recall that $\frac{d}{dx} \log(ax) = \frac{1}{x}$ (the constant $a$ disappears in the derivative of a log). So:

3. Integrate $dv$ to get $v$

(We don’t need the constant of integration yet — it will appear at the end.)

4. Apply the integration by parts formula

Simplify the integral on the right:

5. Evaluate the remaining integral

So: …

Method: Integration by parts with $\log(ax)$ — the constant inside doesn't change $du$

Identical by-parts strategy as $x\log x$; the only subtlety is a constant $a$ inside the logarithm, which vanishes on differentiation.

Steps

Step 1: Choose $u=\log(ax)$ (ILATE: log before algebraic).

Set $u=\log(ax)$, $dv=x\,dx$.

Step 2: Differentiate the log — the inside constant disappears.

Since $\log(ax)=\log a+\log x$ and $\log a$ is constant,

exactly as for $\log x$. So $a$ has no effect on $du$.

Step 3: Integrate $dv$ and apply the formula. …

Common Mistakes

Mistake 1: Thinking the constant inside $\log(2x)$ changes $du$.

Why it's wrong: $\log(2x)=\log 2+\log x$, so $\frac{d}{dx}\log(2x)=\frac1x$ — the same as for $\log x$. Correct approach: Use $du=\frac1x\,dx$; the $2$ affects nothing in the derivative.

Mistake 2: Choosing the algebraic factor as $u$.

Why it's wrong: Taking $u=x$ means integrating $\log(2x)$, which is harder. Correct approach: By ILATE, $u=\log(2x)$, $dv=x\,dx$. …