# Integral of x ln x: Formula, Steps and Examples

> The integral of x ln x is (x²/2) ln x − x²/4 + C. See the integration by parts steps with u = ln x, a check, a graph and the xⁿ ln x formula.

Source: https://practicecalculus.com/integrals/x-ln-x
Updated: 2026-11-06

$$ \int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C $$

## At a glance

- **Antiderivative:** $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$
- **Method:** Integration by parts with $u = \ln x$, $dv = x\,dx$
- **From 1 to e:** $\frac{e^2 + 1}{4} \approx 2.097$
- **From 0 to 1:** $-\frac{1}{4}$
- **General power:** $\int x^n\ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C$, $n \ne -1$

## Key takeaways

- The integral of $x\ln x$ is $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$.
- Pick $u = \ln x$, not $u = x$. You cannot integrate $\ln x$ easily, but differentiating it gives the simple $\frac{1}{x}$.
- The same method works for any power: $\int x^n\ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C$ for $n \ne -1$.
- For $n = -1$, the integral $\int\frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2} + C$ is a substitution instead.

The **integral of $x\ln x$ is $\dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$**. It is a product of a power and a log, so integration by parts is the tool, with the log as $u$.

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

## Graph of x ln x with the area from 1 to e

```graph
title: 'y = x ln x with the area from 1 to e shaded'
caption: 'The shaded area is $\frac{e^2 + 1}{4} \approx 2.097$. Between $0$ and $1$ the curve dips below the axis, and that small region has signed area $-\frac{1}{4}$.'
x: [-0.5, 3.5]
y: [-1, 4.5]
functions:
  - expr: x*ln(x)
    label: y = x ln x
    domain: [0.005, 3.5]
shade:
  expr: x*ln(x)
  from: 1
  to: 2.71828
points:
  - [1, 0, '(1, 0)']
  - [2.71828, 2.71828, '(e, e)']
```

## How to integrate x ln x step by step

Use [integration by parts](/rules/integration-by-parts-rule): $\int u\,dv = uv - \int v\,du$.

1. **Choose the parts.** $u = \ln x$ and $dv = x\,dx$. Logs come first in LIATE.
2. **Find $du$ and $v$.** $du = \frac{1}{x}\,dx$ and $v = \frac{x^2}{2}$.
3. **Apply the formula.**
$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int\frac{x^2}{2}\cdot\frac{1}{x}\,dx = \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx$$
4. **Finish.** $\frac{1}{2}\int x\,dx = \frac{x^2}{4}$, so
$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

The key moment is step 3: the $\frac{1}{x}$ from $du$ cancels one power of $x$ from $v$, leaving a plain power. That always happens with a log as $u$.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{x^2}{2}\ln x - \frac{x^2}{4}\right) = x\ln x + \frac{x^2}{2}\cdot\frac{1}{x} - \frac{x}{2} = x\ln x + \frac{x}{2} - \frac{x}{2} = x\ln x$$

The [derivative of x ln x](/derivatives/x-ln-x) page has the forward calculation.

## Definite integrals

**From 1 to e:**

$$\left[\frac{x^2}{2}\ln x - \frac{x^2}{4}\right]_1^e = \left(\frac{e^2}{2} - \frac{e^2}{4}\right) - \left(0 - \frac{1}{4}\right) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2+1}{4}$$

That is about $2.097$.

**From 0 to 1:** this is [improper](/blog/improper-integrals) at $0$ because $\ln x$ is undefined there, but $x^2\ln x \to 0$ as $x \to 0^+$:

$$\int_0^1 x\ln x\,dx = \left(0 - \frac{1}{4}\right) - 0 = -\frac{1}{4}$$

## The general power xⁿ ln x

The same steps with $dv = x^n\,dx$ and $v = \frac{x^{n+1}}{n+1}$ give

$$\int x^n\ln x\,dx = \frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2} + C, \qquad n \ne -1$$

| Integral | Answer |
| --- | --- |
| $\int\ln x\,dx$ ($n = 0$) | $x\ln x - x + C$ |
| $\int x^2\ln x\,dx$ | $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$ |
| $\int\sqrt{x}\ln x\,dx$ | $\frac{2}{3}x^{3/2}\ln x - \frac{4}{9}x^{3/2} + C$ |
| $\int\frac{\ln x}{x^2}\,dx$ ($n = -2$) | $-\frac{\ln x}{x} - \frac{1}{x} + C$ |

## Worked examples

**Example 1: a cube.** $\int x^3\ln x\,dx = \frac{x^4}{4}\ln x - \frac{x^4}{16} + C$, from the general formula with $n = 3$.

**Example 2: a log of a power.** $\int x\ln(x^2)\,dx$. Use $\ln(x^2) = 2\ln x$ for $x > 0$:

$$2\int x\ln x\,dx = x^2\ln x - \frac{x^2}{2} + C$$

**Example 3: the n = −1 case.** $\int\frac{\ln x}{x}\,dx$ breaks the formula (it would divide by zero). Use $u = \ln x$, $du = \frac{dx}{x}$ instead: $\int u\,du = \frac{(\ln x)^2}{2} + C$.

**Example 4: a squared log.** $\int x(\ln x)^2\,dx$. Parts with $u = (\ln x)^2$, $dv = x\,dx$ gives $\frac{x^2}{2}(\ln x)^2 - \int x\ln x\,dx$, which you now know:

$$\frac{x^2}{2}(\ln x)^2 - \frac{x^2}{2}\ln x + \frac{x^2}{4} + C$$

## Common mistakes

1. **Choosing $u = x$.** Then $v$ needs $\int\ln x\,dx$, and the leftover integral is harder than the original.
2. **Forgetting to simplify $\frac{x^2}{2}\cdot\frac{1}{x}$.** It is $\frac{x}{2}$, and that integrates to $\frac{x^2}{4}$, not $\frac{x^2}{2}$.
3. **Integrating the factors separately.** $\frac{x^2}{2}(x\ln x - x)$ is not the answer.
4. **Using the general formula with $n = -1$.** See Example 3.

Related pages: the [integral of ln x](/integrals/ln-x), the [integral of x eˣ](/integrals/x-e-to-the-x) and the [integration by parts guide](/blog/integration-by-parts).

## Frequently asked questions

### What is the integral of x ln x?

It is $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$, which can also be written $\frac{x^2}{4}(2\ln x - 1) + C$.

### What is the integral of x² ln x?

It is $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$. Use parts with $u = \ln x$ and $dv = x^2\,dx$.

### What is the integral of x ln x from 0 to 1?

It is $-\frac{1}{4}$. At the lower limit, $\frac{x^2}{2}\ln x \to 0$ as $x \to 0^+$, so only $-\frac{1}{4}$ from the upper limit remains.

### Why choose u = ln x and not u = x?

With $u = x$ you would need $v = \int\ln x\,dx = x\ln x - x$, and the leftover integral gets messier. With $u = \ln x$, $du = \frac{1}{x}dx$ cancels a power of $x$ and leaves a plain power to integrate.