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Chapter 5: Q 88. (page 431)

Use the product rule to derive the formula for integration by parts in theorem 5.7.

Short Answer

Expert verified

The solution is $\int u v d x = u \int v d x - \int [\frac{d}{d x} (u) \int v d x] d x$ .

Step by step solution

01

Step 1. Given information.

The product rule of differentiation.

$\frac{d}{d x} (u v) = u \frac{d}{d x} (v) + v \frac{d}{d x} (u)$

02

Step 2. Derivation of the formula for integration by parts.

If $y = u v$ then, $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ .

Rearranging this rule:

$u \frac{d v}{d x} = \frac{d (u v)}{d x} - v \frac{d u}{d x}$

Now integrate both sides:

$\int u \frac{d v}{d x} d x = \int \frac{d (u v)}{d x} d x - \int v \frac{d u}{d x} d x$

The first term on the right simplifies since we are simply integrating.

$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$

03

Step 3. Simplified answer.

This is the formula known as integration by parts.

$\int u \frac{d v}{d x} d x = u v - \int v \frac{d u}{d x} d x$

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Most popular questions from this chapter

Complete the square for each quadratic in Exercises 28–33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x^{2} + 6 x - 2$

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int x \sqrt{x^{2} + 1} d x$

Complete the square for each quadratic in Exercises 28–33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x^{2} - 4 x - 8$

Solve the integral: $\int x \ln (x^{2}) d x$

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}} d x$

See all solutions

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