Q.Evaluate the definite integral: $\int_2^3 \frac{1}{x} \ dx$

The key idea is that the antiderivative of $\frac{1}{x}$ is $\log|x|$, so this is a direct application of the natural logarithm integration rule.

Step 1: Write the integral in standard form:

Step 2: Apply the fundamental theorem of calculus. The antiderivative of $\frac{1}{x}$ is $\log|x|$:

Step 3: Evaluate at the limits:

Step 4: Use the logarithm property $\log a - \log b = \log\left(\frac{a}{b}\right)$:

The core idea here is the fundamental relationship between the natural logarithm and the integral of $1/x$. When you see $\int \frac{1}{x} \, dx$, your mind should immediately jump to $\log|x| + C$, because the derivative of $\log x$ is $1/x$ (for $x > 0$). This is not a coincidence — it’s the definition of the natural logarithm for many mathematicians: $\log a = \int_1^a \frac{1}{t} \, dt$.

Since our limits of integration are from $2$ to $3$, both positive, we can safely drop the absolute value and work directly with $\log x$.

1. Set up the antiderivative.

   The indefinite integral is $\int \frac{1}{x} \, dx = \log x + C$. For a definite integral, we don’t need the constant — we evaluate the antiderivative at the upper and lower limits.

2. Apply the Fundamental Theorem of Calculus.

   This theorem tells us:

where $F$ is any antiderivative of $f$. Here, $F(x) = \log x$, so:

3. Simplify using logarithm properties. The difference of two logs with the same base is the log of the quotient:

Method: Definite integral of $\tfrac1x$ via the natural logarithm

Recognise $\int\dfrac1x\,dx=\log|x|$, then apply the Fundamental Theorem across the limits.

Steps

Step 1: Recall the antiderivative. $\int\dfrac{1}{x}\,dx=\log|x|+C$ (natural log).

Step 2: Set up the bracket.

Step 3: Use the log-subtraction law.

$\log b-\log a=\log\frac{b}{a},$ giving a single compact logarithm.

Step 4: Report the exact value (e.g. $\log\tfrac32$); keep it symbolic unless a decimal is asked.

Common Mistakes

Mistake 1: Writing $\int\dfrac1x\,dx=-\dfrac{1}{x^{2}}$ (misapplying the power rule).

Why it's wrong: the power rule fails for $n=-1$; the antiderivative is $\log|x|$. Correct approach: $\int x^{-1}dx=\log|x|+C$.

Mistake 2: Turning $\log b-\log a$ into $\dfrac{\log b}{\log a}$.

Why it's wrong: subtraction of logs is division of arguments, not of logs. Correct approach: $\log b-\log a=\log\dfrac{b}{a}$.

Mistake 3: Evaluating as $\log(b-a)$.

Why it's wrong: $\log 3-\log 2\neq\log 1$. Correct approach: use $\log\dfrac32$.