Math · Calculus II · Worked example

Integrate ln x by parts

Find the integral of ln x, then evaluate it from x = 1 to x = e.

\int ln x d x

Choose u and dv

There is no obvious product, but ln x is easy to differentiate and dx is easy to integrate.

The parts for the integral of ln x
Part	Choice	Then
u	ln x	du = dx/x
dv	dx	v = x

The new integrand, x · (1/x), is just 1.

\int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x = x ln x - x + C

\begin{array}{c} \int ln x d x = x ln x - \int x \cdot \frac{1}{x} d x \\ = x ln x - x + C \end{array}

With F(x) = x ln x − x, F(e) − F(1) = (e − e) − (0 − 1) = 1.

\begin{array}{l} F (x) = x ln x - x \\ F^{'} (x) = ln x \\ F (e) - F (1) = 1 \end{array}

x ln x − x + C; from 1 to e the integral is 1.

Find the integral of x ln x.

Show the answer and explanation

Take u = ln x, which simplifies when differentiated, and dv = x dx, so du = dx/x and v = x²/2. The new integral is of x/2, which gives x²/4.

\begin{array}{l} \int x ln x d x \\ \frac{x^{2}}{2} ln x - \frac{x^{2}}{4} + C \end{array}

In Derivative & antiderivative checks, enter x ln x alone. Its derivative is ln x + 1, so the checker rejects it: the − x is what cancels the 1.

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