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

The key idea is to treat this as a product of two functions — $x$ and $\log x$ — and apply integration by parts.

We set:

• $u = \log x$ (so $du = \frac{1}{x} dx$)
• $dv = x \, dx$ (so $v = \frac{x^2}{2}$)

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

Simplify the remaining integral: …

Why integration by parts?

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

The trick is choosing which part to call $u$ and which to call $dv$. For products involving $\log x$, a reliable rule of thumb is: let $u$ be the logarithmic function, because its derivative is simpler ($\frac{1}{x}$), while the polynomial part becomes easy to integrate repeatedly.

If we instead set $u = x$ and $dv = \log x \, dx$, we'd need to integrate $\log x$ — which is doable, but then $v$ would be $x \log x - x$, and the resulting integral becomes messier. The first choice is cleaner.

Step-by-step solution

1. Set up the parts.

Let

Then differentiate $u$ and integrate $dv$:

2. Apply the integration by parts formula.

3. Simplify the new integral.

The $x$ cancels:

So we have:

4. Integrate the remaining term. …

Method: Integration by parts with a logarithm — log is always $u$

For a product involving $\log x$, integration by parts is the tool, and the logarithm is always the $u$ (it has no elementary "nice" antiderivative but a very simple derivative).

Steps

Step 1: Choose $u=\log(\ldots)$ by ILATE.

Logarithmic beats Algebraic, so $u=\log x$ and $dv=x\,dx$. Differentiating the log turns it into a rational function $\dfrac1x$, which cancels nicely against $v$.

Step 2: Compute $du$ and $v$.

Step 3: Apply the formula. …

Common Mistakes

Mistake 1: Choosing $u=x$, $dv=\log x\,dx$.

Why it's wrong: This forces you to integrate $\log x$ first ($v=x\log x-x$), making the problem messier. Correct approach: By ILATE the logarithm is $u$; then $du=\frac1x\,dx$ is simple and cancels against $v$.

Mistake 2: Sign error in $uv-\int v\,du$.

Why it's wrong: The formula subtracts the second integral; writing $+\int v\,du$ flips the $-\frac{x^2}{4}$ term. Correct approach: Apply $\int u\,dv=uv-\int v\,du$ exactly. …