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Q 41.

Expert-verified

Calculus

Found in: Page 429

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Solve the integral: $\int \frac{x^{2} + 1}{e^{x}} d x$

The required answer is $- \frac{2 x}{e^{x}} - \frac{x^{2} + 1}{e^{x}} - 2 e^{- x} + C$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information.

We have given integral is $\int \frac{x^{2} + 1}{e^{x}} d x$ .

Step 2. Solve the integration by parts .

We have,

$u = x^{2} + 1 d u = 2 x d x$

and

role="math" localid="1648720139610" $d v = e^{- x} d x v = \int e^{- x} d x v = - \frac{1}{e^{x}}$

The formula of integration by parts is $\int u d v = u v - \int v d u$ .

role="math" localid="1648720506794" $\int \frac{x^{2} + 1}{e^{x}} d x = [(x^{2} + 1) (- \frac{1}{e^{x}}) - \int (- \frac{1}{e^{x}}) d x] = [- \frac{1}{e^{x}} (x^{2} + 1) - \int (- \frac{2 x}{e^{x}}) d x] = [- \frac{1}{e^{x}} (x^{2} + 1) + 2 \int (\frac{x}{e^{x}}) d x]$

Step 3. Integration by parts.

We have, $u = x d u = d x$

and $d v = e^{- x} d x v = \int e^{- x} d x v = - \frac{1}{e^{x}}$

$= = [- \frac{1}{e^{x}} (x^{2} + 1) + 2 \int (e^{- x}) d x] = [- \frac{1}{e^{x}} (x^{2} + 1) + 2 (x (- \frac{1}{e^{x}}) - \int (- \frac{1}{e^{x}}) d x)] = [- \frac{1}{e^{x}} (x^{2} + 1) + 2 (- \frac{x}{e^{x}} - \int (- \frac{1}{e^{x}}) d x)] = [- \frac{1}{e^{x}} (x^{2} + 1) + 2 (- \frac{x}{e^{x}} - (+ e^{- x}) d x)] = - \frac{2 x}{e^{x}} - \frac{1}{e^{x}} (x^{2} + 1) - 2 e^{- x} + c$

Most popular questions for Math Textbooks

Find three integrals in Exercises 27–70 for which a good strategy is to apply integration by parts twice.

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.

$\int \frac{\sqrt{9 - x^{2}}}{x} d x$

Find three integrals in Exercises 27–70 for which a good strategy is to use integration by parts with $u = x$ and dv the remaining part.

Explain why, if $x = a \sin u$ , then $\sqrt{a^{2} - x^{2}} = a \cos u$ . Your explanation should include a discussion of domains and absolute values.

Why don’t we need to have a square root involved in order to apply trigonometric substitution with the tangent? In other words, why can we use the substitution $x = a \tan u$ when we see $x^{2} + a^{2}$ , even though we can’t use the substitution $x = a \sin u$ unless the integrand involves the square root of $a^{2} - x^{2}$ ? (Hint: Think about domains.)

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