Q.Integrate the following function: $\frac{(1 + \log x)^2}{x}$

The key idea is U Substitution: set $u = 1 + \log x$, because its derivative $\frac{1}{x}$ appears in the integrand.

Step 1: Let $u = 1 + \log x$. Then $du = \frac{1}{x} dx$.

Step 2: The integral becomes

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

Step 3: Integrate: $\int u^2 \, du = \frac{u^3}{3} + C$. …

Why substitution works here

When you see a function like $\frac{(1 + \log x)^2}{x}$, the natural instinct might be to expand the square. But that would lead to three separate terms, each needing its own integration — messy and unnecessary.

Instead, notice the structure: the numerator contains $(1 + \log x)^2$, and the denominator is $x$. The derivative of $\log x$ is $\frac{1}{x}$, which means the derivative of $1 + \log x$ is also $\frac{1}{x}$. That $\frac{1}{x}$ is sitting right there in the integrand, waiting to pair with a substitution.

This is the classic pattern for u-substitution: you have a composite function (something squared) multiplied by the derivative of its inner part. The substitution collapses the whole expression into a simple power.

Step-by-step solution

1. Choose the substitution

   Let $u = 1 + \log x$.

   Why this? Because the integrand has $(1 + \log x)^2$, and we suspect its derivative will appear.

2. Differentiate to find $du$

   $\frac{du}{dx} = \frac{1}{x}$, so $du = \frac{1}{x} \, dx$.

   Notice that $\frac{dx}{x}$ is exactly the factor that multiplies $(1 + \log x)^2$ in the original integral.

3. Rewrite the integral in terms of $u$

   The original integral is $\int \frac{(1 + \log x)^2}{x} \, dx = \int (1 + \log x)^2 \cdot \frac{1}{x} \, dx$.

   Substituting $u$ and $du$ gives:

4. Integrate with respect to $u$ This is a standard power rule: …

Method: Substitution driven by $\dfrac{d}{dx}(\log x)=\dfrac1x$

Use this whenever the integrand is some function of $\log x$ multiplied by $\dfrac1x$ — the $\dfrac1x$ is exactly the derivative of $\log x$, which makes a substitution collapse the integral.

Steps

Step 1: Spot the $\dfrac1x$ factor.

Any $\dfrac1x$ in the integrand should be read as "the derivative of $\log x$ is sitting here." That is the signal to substitute.

Step 2: Let $u$ equal the whole expression built from $\log x$.

Choose $u$ to be the inner block that is being raised to a power or fed into a function — here that is $1+\log x$, not just $\log x$, so the algebra stays clean:

$u = 1+\log x,\qquad du=\frac1x\,dx$ …

Common Mistakes

Mistake 1: Expanding $(1+\log x)^2$ before integrating.

Why it's wrong: expanding gives $1+2\log x+(\log x)^2$ over $x$, three harder integrals (one needing integration by parts), when the structure already invites a one-line substitution. Correct approach: set $u=1+\log x$ so $du=\tfrac1x\,dx$ and integrate $\int u^2\,du$.

Mistake 2: Choosing $u=\log x$ and mishandling the constant $1$. …