1.8. Integration by parts

Integrals like

(1.41)\[\begin{equation} \int x e^{-6 x} \d{x} \end{equation}\]

are not solvable by substitution, but they look like they are related closely enough that you may be able to! What to do here? Recall the product rule for two functions u and v, then integrate:

(1.42)\[\begin{align} (uv)' &= uv' + v u' \\ uv &= \int u \d{v} + \int v \d{u} \end{align}\]

Rearranging gives the familiar form:

Integration by parts

(1.43)\[\begin{equation} \int u \d{v} = uv - \int v \d{u} \end{equation}\]

To apply this technique, identify a u and a \(\d{v}\). Differentiate \(u\) and integrate \(\d{v}\), then apply the formula.

Let’s try this for the integral above with

(1.44)\[\begin{align} u &= x & \d{v} &= e^{-6x} \d{x} \\ \d{u} &= \d{x} & v &= -\frac{e^{-6x}}{6} \end{align}\]

so

(1.45)\[\begin{align} \int x e^{-6x} \d{x} &= x\frac{-e^{-6x}}{6} - \int\frac{-1}{6} e^{-6x} \d{x} \\ &= -\frac{xe^{-6x}}{6} + \frac{1}{6} \int e^{-6x} \d{x} \\ &= -\frac{xe^{-6x}}{6} - \frac{e^{-6x}}{36} + c \end{align}\]

1.8.1. Choosing parts

The heart of the technique is making a suitable choice for u and \(\d{v}\). The acronym L.I.A.T.E. can help make this choice:

• Logarithmic (best choice of u)

• Inverse

• Algebraic

• Trigonometric

• Exponential (worst choice of u)

Example: Integration by parts 1

Evaluate

(1.46)\[\begin{equation} \int(3x+5)\cos\left(\frac{x}{4}\right) \d{x} \end{equation}\]

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Let

(1.47)\[\begin{align} u &= 3x+5 & \d{v} &= \cos\left(\frac{x}{4}\right) \d{x} \\ \d{u} &= 3 \d{x} & v &= 4\sin\left(\frac{x}{4}\right) \end{align}\]

so:

(1.48)\[\begin{align} \int(3x+5)\cos\left(\frac{x}{4}\right) \d{x} &= (3x+5)\left[4\sin\left(\frac{x}{4}\right)\right] - \int 4 \sin \left(\frac{x}{4}\right) 3 \d{x} \\ &= (12x+20) \sin\left(\frac{x}{4}\right) - 12\int\sin\left(\frac{x}{4}\right) \d{x} \\ &= (12x+20) \sin\left(\frac{x}{4}\right) + 48\cos\left(\frac{x}{4}\right) + c \end{align}\]

Example: Integration by parts 2

Evaluate

(1.49)\[\begin{equation} \int \ln x \d{x} \end{equation}\]

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Let

(1.50)\[\begin{align} u &= \ln x & \d{v} &= \d{x} \\ \d{u} &= \frac{1}{x} \d{x} & v &= x \end{align}\]

so:

(1.51)\[\begin{align} \int \ln x \d{x} &= x \ln x - \int\frac{1}{x} x \d{x} \\ &= x \ln x - \int \d{x} \\ &= x \ln x - x + c \\ \end{align}\]

Example: Integration by parts 3

Evaluate

(1.52)\[\begin{equation} \int x^5 \sqrt{x^3+1} \d{x} \end{equation}\]

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Let

(1.53)\[\begin{align} u &= x^3 & \d{v} &= x^2\sqrt{x^3+1} \d{x} \\ \d{u} &= 3x^2 \d{x} & v &= \frac{2}{9}(x^3+1)^{3/2} \end{align}\]

so:

(1.54)\[\begin{align} \int(x^5)(\sqrt(x^3+1))dx &= x^3\frac{2}{9}(x^3+1)^{3/2} - \int\left[\frac{2}{9}(x^3+1)^{3/2}\right]3x^2dx \\ &= \frac{2}{9}x^3(x^3+1)^{3/2} - \frac{2}{9}\cdot\frac{2}{5}(x^3+1)^{5/2} + c \\ &= \frac{2}{9}x^3(x^3+1)^{3/2} - \frac{4}{45}(x^3+1)^{5/2} + c \end{align}\]

1.8.2. Tabular method

To integrate

(1.55)\[\begin{equation} \int x^4 e^{x/2} \d{x} \end{equation}\]

by parts, let

(1.56)\[\begin{align} u &= x^4 & \d{v} &= e^{x/2} \\ \d{u} &= 4x^3 \d{x} & v &= 2e^{x/2} \end{align}\]

which allows:

(1.57)\[\begin{equation} \int x^4e^{x/2} \d{x} = 2 x^4 e^{x/2} - \int 8 x^3 e^{x/2} \d{x} \end{equation}\]

We will need to integrate by parts again, and again, and again… but there is a shortcut! Make a staggered table. Differentiate u repeatedly until you get a 0 in the last row. Then, integrate \(\d{v}\) repeatedly until you reach the last nonzero row for u. Multiply across rows using opposite signs.

sign	\(u\)	\(\d{v}\)
		\(e^{x/2}\)
\(+\)	\(x^4\)	\(2e^{x/2}\)
\(-\)	\(4x^3\)	\(4e^{x/2}\)
\(+\)	\(12x^2\)	\(8e^{x/2}\)
\(-\)	\(24x\)	\(16e^{x/2}\)
\(+\)	\(24\)	\(32e^{x/2}\)
	\(0\)

The result is:

(1.58)\[\begin{align} \int x^4 e^{x/2} \d{x} &= x^4 (2e^{x/2}) - 4x^3 (4e^{x/2}) \\ &+ 12x^2 (8e^{x/2}) - 24x (16e^{x/2}) + 24 (32e^{x/2}) \\ &= (2x^4 - 16x^3 + 96x^2 - 384x + 768) e^{x/2} \end{align}\]