Integral of ln x: Formula, Steps and Examples

The integral of \ln x is x\ln x - x + C. There is no reverse rule that undoes the log directly, so the standard method is integration by parts with a hidden factor of 1.

Graph of ln x and its antiderivative

How to integrate ln x step by step

Use integration by parts, \int u\,dv = uv - \int v\,du.

1. Write \ln x as a product. \int\ln x\,dx = \int \ln x\cdot 1\,dx.
2. Choose the parts. u = \ln x and dv = dx. Then du = \frac{1}{x}\,dx and v = x.
3. Apply the formula. \int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx
4. Finish. \int 1\,dx = x, so the answer is x\ln x - x + C.

Why does this work? Differentiating \ln x turns it into \frac{1}{x}, which cancels the x from v. In the LIATE order, logs come first, so they almost always play the role of u. The integration by parts guide has more on choosing u.

Check by differentiating

By the product rule and the derivative of ln x:

Definite integrals

From 1 to e:

From 1 to 2: (2\ln 2 - 2) - (0 - 1) = 2\ln 2 - 1 \approx 0.386.

From 0 to 1: this is an improper integral since \ln x \to -\infty near 0. Using the limit x\ln x \to 0 as x \to 0^+:

The answer is negative because \ln x < 0 on (0, 1), so the area lies below the axis.

Worked examples

Example 1: a constant inside. \int\ln(2x)\,dx. Let u = 2x, dx = \frac{du}{2}:

Example 2: a shifted log. \int\ln(x+1)\,dx. Let u = x + 1, du = dx:

The -1 can be absorbed into C, so (x+1)\ln(x+1) - x + C is also correct.

Example 3: a squared log. \int(\ln x)^2\,dx. Take u = (\ln x)^2, dv = dx, so du = \frac{2\ln x}{x}\,dx and v = x:

Example 4: a power of x inside. \int\ln(x^3)\,dx = 3\int\ln x\,dx = 3x\ln x - 3x + C, after the log rule \ln(x^3) = 3\ln x.

Example 5: base 10. Since \log_{10}x = \frac{\ln x}{\ln 10}, \int\log_{10}x\,dx = \frac{x\ln x - x}{\ln 10} + C.

Common mistakes

1. Answering \frac{1}{x}. That is the derivative of \ln x. The integral of \frac{1}{x} is \ln\lvert x\rvert; see the integral of 1/x.
2. Choosing u = 1, dv = \ln x\,dx. Then you need \int\ln x\,dx to find v, the very thing you are trying to compute.
3. Losing the -x. x\ln x alone has derivative \ln x + 1, one too many.
4. Confusing \ln x with \frac{\ln x}{x}. \int\frac{\ln x}{x}\,dx = \frac{(\ln x)^2}{2} + C is a u-substitution, not a parts problem.

For a log times a power of x, see the integral of x ln x. For logs in the denominator, see the integral of 1/(x ln x).