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

The key idea is U Substitution — noticing that the derivative of $\log x$ is $\frac{1}{x}$, which appears in the integrand.

Let $u = \log x$. Then $du = \frac{1}{x} dx$, so the integral becomes:

Integrate: $\int u^2 \, du = \frac{u^3}{3} + C$. Substitute back $u = \log x$:

Why U-Substitution Works Here

When you see a function like $\frac{(\log x)^2}{x}$, the key is to notice that the derivative of $\log x$ is $\frac{1}{x}$, which is already sitting in the denominator. This is the classic signal for substitution: if you let $u$ be the "inside" function whose derivative appears nearby, the integral collapses into something much simpler.

Think of it as untangling a knot. The expression $(\log x)^2$ is the complicated part, and $\frac{1}{x}$ is the tool that helps you straighten it out. By setting $u = \log x$, you replace the messy $\log x$ with a clean variable $u$, and the $\frac{1}{x} \, dx$ becomes $du$. Suddenly, you're just integrating $u^2$, which is as straightforward as it gets.

Step-by-Step Solution

1. Set up the substitution. Let $u = \log x$. Then differentiate:

This is the crucial link — the $dx$ in the integral pairs with the $\frac{1}{x}$ to form $du$.

2. Rewrite the integral in terms of $u$. The original integral is

Substituting $u = \log x$ and $du = \frac{1}{x} dx$ gives:

3. Integrate using the power rule. The power rule for integrals says $\int u^n \, du = \frac{u^{n+1}}{n+1} + C$ for $n \neq -1$. Here $n = 2$, so:

4. Substitute back to the original variable. Recall $u = \log x$, so:

Method: Substitute the inner function when its derivative is also present

Use this when the integrand contains a function of $\log x$ (or any inner function) multiplied by that inner function's derivative — here $(\log x)^2$ times $\dfrac1x$.

Steps

Step 1: Spot the inner function and its derivative.

$\dfrac{d}{dx}(\log x)=\dfrac1x$, and the $\dfrac1x$ factor is already sitting in the integrand — a signal to substitute.

Step 2: Let $u$ be the inner function.

Put $u=\log x$, so $du=\dfrac{1}{x}\,dx$. The integral collapses to a pure power of $u$: $\displaystyle\int u^2\,du$.

Step 3: Integrate in $u$, then back-substitute.

The whole method rests on recognising a "$\text{function of }g(x)$ times $g'(x)$" shape.

Common Mistakes

Mistake 1: Not spotting the $\dfrac1x$ as $d(\log x)$.

Why it's wrong: without the substitution $u=\log x$, students try to integrate $(\log x)^2$ directly, which has no elementary power-rule form here. Correct approach: since $du=\dfrac1x\,dx$ is present, substitute and integrate $u^2$.

Mistake 2: Applying the power rule to $\log x$ as if it were $x$.

Why it's wrong: $\int(\log x)^2\,dx\neq\dfrac{(\log x)^3}{3}$ on its own — the $\dfrac1x$ factor is what makes it valid. Correct approach: the result $\dfrac{(\log x)^3}{3}+C$ holds only because $\dfrac1x\,dx=du$.

Mistake 3: Dropping the $+C$.

Why it's wrong: indefinite integrals need the constant. Correct approach: finish with $+C$.