Chapter 5: Q. 42 (page 407)

Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.
$\int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x$

Short Answer

Expert verified

The value is $\infty$ .

Step by step solution

Step 1. Given information.

The given function is $\int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x$ .

Step 2. Value of the integral.

The given integral can be written as,

$\int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x = \lim_{B \to \infty} \int_{0}^{B} \frac{x}{(x^{2} + 1)} d x$

$Let u = x^{2} + 1 i.e d u = 2 x d x T h e r e f o r e, \int \frac{x}{(x^{2} + 1)} d x = \int \frac{1}{2 u} d u = \frac{1}{2} \int \frac{1}{u} d u = \frac{1}{2} \ln | u | \int \frac{x}{(x^{2} + 1)} d x = \frac{1}{2} \ln |x^{2} + 1|$

Step 3. Substitution.

Substituting the obtained value in the given equation, we get,

$\int \frac{x}{(x^{2} + 1)} d x = \frac{1}{2} \ln |x^{2} + 1| \int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x = \lim_{B \to \infty} \frac{1}{2} \ln |x^{2} + 1| = \infty - 0 = \infty$

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Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.
$\int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of the integral.

Step 3. Substitution.

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Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.∫0∞xx2+1dx

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of the integral.

Step 3. Substitution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Theoretical and Mathematical Physics

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Geometry

Study anywhere. Anytime. Across all devices.

Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.
$\int_{0}^{\infty} \frac{x}{(x^{2} + 1)} d x$