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Products and integration by parts

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.

Products and integration by parts Updated 3 min read

At a glance

\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C
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
On this page
  1. Graph of x ln x with the area from 1 to e
  2. How to integrate x ln x step by step
  3. Check by differentiating
  4. Definite integrals
  5. The general power xⁿ ln x
  6. Worked examples
  7. Common mistakes

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

y = x ln x with the area from 1 to e shaded
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}.

How to integrate x ln x step by step

Use integration by parts: \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 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 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, the integral of x eˣ and the integration by parts guide.

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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.

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Related integrals

Integral of ln x \int \ln x\,dx = x\ln x - x + C The integral of ln x is x ln x − x + C. See the integration by parts steps, a check by differentiating, a graph and examples like ln(2x) and (ln x)². Integral of x eˣ \int x e^x\,dx = (x - 1)e^x + C The integral of x eˣ is (x − 1)eˣ + C. See the integration by parts steps, why u = x is the right choice, a graph and examples like x e^(2x). Integral of 1/(x ln x) \int \frac{1}{x\ln x}\,dx = \ln\lvert\ln x\rvert + C The integral of 1/(x ln x) is ln|ln x| + C. See the u = ln x substitution, why its improper integral diverges, a graph and worked examples.