The formula, from the product rule

\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}

uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx

\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx

the two integrals are the two areas either side of the curve; here u = ln x and v = x

Integration by parts

One area you can't find, next to one you can.

Integration by parts looks like pure algebra. Draw it and it is a rectangle cut in two by a curve: know one piece and the rectangle gives you the other.

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An area with no rule

The shaded area is $\displaystyle\int_1^3 \ln x\,dx$ . None of the standard integrals is $\ln x$ , so there is no rule to reverse here.

The other side of the curve

The curve cuts the $3 \times \ln 3$ rectangle in two. The other piece, between the curve and the $y$ -axis, is $\int x\,dy$ . Along the curve $dy = \tfrac{1}{x}\,dx$ , so it is

\int_1^3 x \times \frac{1}{x}\,dx = \int_1^3 1\,dx = 2

Take the easy piece away

The two pieces fill the rectangle, so the area we wanted is the rectangle minus the easy piece:

\int_1^3 \ln x\,dx = 3\ln 3 - 2

That is $1.2958\ldots$ , the value the trapezium rule estimated in Chapter 14.

Any product

The same trade works for any product. Choose $u$ to be the factor that gets simpler when you differentiate it, like the $x$ in $\int xe^{2x}\,dx$ . $\ln x$ is the exception: it has no standard integral, so it always goes in as $u$ .

Chapter 15

Swap the area you can't find for the one you can.

Standard integrals, reversing the chain rule, substitution, partial fractions, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›