1.8. Integration by parts

Integrals like

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

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:

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

Rearranging gives the familiar form:

Integration by parts

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

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

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

so

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

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

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

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Let

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

so:

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

Example: Integration by parts 2

Evaluate

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

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Let

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

so:

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

Example: Integration by parts 3

Evaluate

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

---

Let

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

so:

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

1.8.2. Tabular method

To integrate

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

by parts, let

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

which allows:

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

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:

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

1.8.3. Skill builder problems

Evaluate the following

1. \(\displaystyle \int (x+2)\ln x \d{x}\)

   Solution

   Select the following parts:

   (1.184)\[\begin{align} u &= \ln x & \d{v} &= (x+2)\d{x} \\ \d{u} &= \frac{1}{x}\d{x} & v &= \frac{x^2}{2} + 2x \end{align}\]

   so

   (1.185)\[\begin{align} \int (x+2)\ln x \d{x} &= \left( \frac{x^2}{2} + 2x \right)\ln x - \int \left( \frac{x^2}{2} + 2x \right) \frac{1}{x} \d{x} \\ &= \left( \frac{x^2}{2} + 2x \right)\ln x - \int \left( \frac{x}{2} + 2 \right) \d{x} \\ &= \left( \frac{x^2}{2} + 2x \right)\ln x - \frac{x^2}{4} - 2x + c \end{align}\]

2. \(\displaystyle \int x^5\sin x \d{x}\)

   Solution

   Since \(x^5\) is to the fifth order, integration by parts must be used 5 times. Use the tabular method for this.

   sign	\(u\)	\(\d{v}\)
   		\(\sin x\)
   \(+\)	\(x^5\)	\(-\cos x\)
   \(-\)	\(5x^4\)	\(-\sin x\)
   \(+\)	\(20x^3\)	\(\cos x\)
   \(-\)	\(60x^2\)	\(\sin x\)
   \(+\)	\(120x\)	\(-\cos x\)
   \(-\)	\(120\)	\(-\sin x\)
   	\(0\)

   so

   (1.186)\[\begin{equation} \int x^5\sin x \d{x} = -x^5 \cos x + 5x^4 \sin x + 20x^3 \cos x - 60x^2 \sin x - 120x \cos x + 120 \sin x + c \end{equation}\]

   To make the solution look cleaner, you can factor out \(\cos x\) and \(\ sin x\).

   (1.187)\[\begin{equation} \int x^5\sin x \d{x} = (-x^5 + 20x^3 - 120x) \cos x + (5x^4 - 60x^2 + 120) \sin x + c \end{equation}\]

3. \(\displaystyle \int e^x\cos x \d{x}\)

   Solution

   Select the following parts:

   (1.188)\[\begin{align} u &= \cos x & \d{v} &= e^x\d{x} \\ \d{u} &= -\sin x\d{x} & v &= e^x \end{align}\]

   so

   (1.189)\[\begin{equation} \int e^x \cos x \d{x} = e^x \cos x + \int e^x \sin x \d{x} \end{equation}\]

   To evaluate this integral, select new parts:

   (1.190)\[\begin{align} u &= \sin x, & \d{v} &= e^x\d{x} \\ \d{u} &= \cos x\d{x} & v &= e^x \end{align}\]

   so

   (1.191)\[\begin{equation} \int e^x \cos x \d{x} = e^x \cos x+ \left[e^x \sin x - \int e^x \cos x \d{x}\right] \end{equation}\]

   The same integral appears on both sides! Add them together and solve:

   (1.192)\[\begin{align} 2 \int e^x \cos x \d{x} &= e^x \cos x + e^x \sin x \\ \int e^x \cos x \d{x} &= \frac{e^x}{2}(\cos x + \sin x) + c \end{align}\]