Find $\int \log x\, dx$

> [!TLDR] > The integral of $\log x$ is solved using integration by parts, treating $\log x$ as $1 \cdot \log x$. The result is $x \log x - x + C$. The key insight here is that $\log x$ doesn't have an obvious antiderivative from the power rule or standard formulas. But we can rewrite it as a product: $\log x = 1 \cdot \log x$. This lets us use **integration by parts**, which is the reverse of the product rule for derivatives. Integration by parts says: $\int u \, dv = uv - \int v \, du$. The trick is choosing $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $\log x$, we set $u = \log x$ because its derivative is $\frac{1}{x}$, a simple rational function. Then $dv = 1 \, dx$, so $v = x$. Let's work through it step by step. 1. **Set up integration by parts.** We have $\int \log x \, dx = \int (1)(\log x) \, dx$. Choose: $u = \log x$ $dv = 1 \, dx$ 2. **Find $du$ and $v$.** Differentiate $u$: $du = \frac{1}{x} \, dx$ Integrate $dv$: $v = \int 1 \, dx = x$ 3. **Apply the integration by parts formula.** $\int u \, dv = uv - \int v \, du$ Substitute: $\int \log x \, dx = (\log x)(x) - \int x \cdot \frac{1}{x} \, dx$ 4. **Simplify the new integral.** $x \cdot \frac{1}{x} = 1$, so we get: $\int \log x \, dx = x \log x - \int 1 \, dx$ 5. **Integrate the remaining term.** $\int 1 \, dx = x + C$ (don't forget the constant of integration) Therefore: $\int \log x \, dx = x \log x - x + C$ > [!WARNING] > A common mistake is to forget the constant of integration $C$ or to misapply the formula by swapping $u$ and $dv$. If you set $u = 1$ and $dv = \log x \, dx$, you'd need to know the integral of $\log x$ already — which is exactly what we're trying to find! Always choose $u$ as the function that simplifies when differentiated. > [!TIP] > This result is a classic and worth memorizing: $\int \log x \, dx = x \log x - x + C$. It also works for $\ln x$ (natural log) —

Find ∫ log x dx

Worked Examples · Example 18

\int \log x\, dx

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Concept understanding — Natural Logarithm Integration

Natural Logarithm Integration: From Intuition to Formula

You already know that integration is the reverse of differentiation. So the first question is: what function, when differentiated, gives $\frac{1}{x}$ ?

You know $\frac{d}{dx}(x^n) = n x^{n-1}$ . Trying to find a function whose derivative is $x^{-1}$ , the power rule would give $\frac{x^0}{0}$ , which is undefined. That's the clue — $\frac{1}{x}$ doesn't fit the power rule pattern.

The function that fills this gap is the natural logarithm, $\log x$ . Its derivative is exactly $\frac{1}{x}$ (for $x > 0$ ), so integration reverses this:

\int \frac{1}{x} \, dx = \log|x| + C

The absolute value $|x|$ is crucial — it extends the formula to negative $x$ , because $\log x$ is only defined for positive numbers, but $\frac{1}{x}$ is defined for all $x \neq 0$ .

Why the absolute value?

For $x > 0$ , differentiating $\log|x|$ gives $\frac{1}{x}$ . For $x < 0$ , $\log|x| = \log(-x)$ , and its derivative is $\frac{1}{-x} \cdot (-1) = \frac{1}{x}$ . Same result, so $\log|x|$ works for both sides.

The generalised form

The real power comes when the numerator is the derivative of the denominator:

\int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| + C

This is the logarithmic integration pattern.

Example to see it in action

Find $\int \frac{2x}{x^2 + 1} \, dx$ . Here $f(x) = x^2 + 1$ , so $f'(x) = 2x$ — the numerator matches. Therefore:

\int \frac{2x}{x^2 + 1} \, dx = \log|x^2 + 1| + C = \log(x^2 + 1) + C

(dropping the absolute value since $x^2 + 1$ is always positive).

What if the numerator doesn't match exactly?

For $\int \frac{x}{x^2 + 1} \, dx$ , the derivative of the denominator is $2x$ but you only have $x$ . Adjust by factoring:

\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int \frac{2x}{x^2 + 1} \, dx = \frac{1}{2} \log|x^2 + 1| + C

Tip

When the numerator is a constant multiple of the derivative of the denominator, factor out that constant: $\int \frac{k \cdot f'(x)}{f(x)} \, dx = k \log|f(x)| + C$ .

Common mistake to avoid

Do not apply this pattern when the numerator is unrelated to the derivative of the denominator. For example, $\int \frac{1}{x^2 + 1} \, dx$ is not $\log|x^2 + 1|$ — it gives $\tan^{-1} x + C$ , a completely different result.

Watch out

The rule only works when the numerator is exactly (or a constant multiple of) the derivative of the denominator. If not, use another method (partial fractions, trigonometric substitution, etc.).

Final formula to remember:

\boxed{\int \frac{f'(x)}{f(x)} \, dx = \log|f(x)| + C}

And the simplest case: $\int \frac{1}{x} \, dx = \log|x| + C$ .

The ∫f'(x)/f(x) dx = log|f(x)| + C pattern is one of the most tested standard results in the NCERT Class 12 Integrals chapter, appearing constantly in CBSE board and JEE Main 'evaluate the integral' questions. Students searching 'integration of 1/x formula' or 'logarithmic integration examples class 12' will find this numerator-matches-derivative-of-denominator rule is exactly the shortcut those exam papers expect students to spot instantly.

The integral of $\log x$ is solved using integration by parts, treating $\log x$ as $1 \cdot \log x$ . The result is $x \log x - x + C$ .

The key insight here is that $\log x$ doesn't have an obvious antiderivative from the power rule or standard formulas. But we can rewrite it as a product: $\log x = 1 \cdot \log x$ . This lets us use integration by parts, which is the reverse of the product rule for derivatives.

Integration by parts says: $\int u \, dv = uv - \int v \, du$ . The trick is choosing $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $\log x$ , we set $u = \log x$ because its derivative is $\frac{1}{x}$ , a simple rational function. Then $dv = 1 \, dx$ , so $v = x$ .

Let's work through it step by step.

Set up integration by parts.

We have $\int \log x \, dx = \int (1)(\log x) \, dx$ .

Choose:

$u = \log x$

$dv = 1 \, dx$
Find $du$ and $v$ .

Differentiate $u$ : $du = \frac{1}{x} \, dx$

Integrate $dv$ : $v = \int 1 \, dx = x$
Apply the integration by parts formula.

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

Substitute:

$\int \log x \, dx = (\log x)(x) - \int x \cdot \frac{1}{x} \, dx$
Simplify the new integral.

$x \cdot \frac{1}{x} = 1$ , so we get:

$\int \log x \, dx = x \log x - \int 1 \, dx$
Integrate the remaining term.

$\int 1 \, dx = x + C$ (don't forget the constant of integration)

Therefore:

$\int \log x \, dx = x \log x - x + C$

Watch out

A common mistake is to forget the constant of integration $C$ or to misapply the formula by swapping $u$ and $dv$ . If you set $u = 1$ and $dv = \log x \, dx$ , you'd need to know the integral of $\log x$ already — which is exactly what we're trying to find! Always choose $u$ as the function that simplifies when differentiated.

Tip

This result is a classic and worth memorizing: $\int \log x \, dx = x \log x - x + C$ . It also works for $\ln x$ (natural log) — same formula. For $\log_a x$ , use the change of base: $\log_a x = \frac{\ln x}{\ln a}$ , then integrate.

✓Final answer

\boxed{x \log x - x + C}

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Q.Find $\int \log x\, dx$

Concept understanding — Natural Logarithm Integration

Natural Logarithm Integration: From Intuition to Formula

Why the absolute value?

The generalised form

Example to see it in action

What if the numerator doesn't match exactly?

Common mistake to avoid

Method: Integration by Parts on a Lone Logarithm (the " $1\cdot$ " Trick)

Steps

Common Mistakes

Q.Find ∫log⁡x dx\int \log x\, dx∫logxdx

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Q.Find $\int \log x\, dx$

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