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Chapter 5: Q. 55 (page 496)

${鈭�}_{0}^{鈭�} x \ln x d x$

Short Answer

Expert verified

Undefined.

The limit does not converge.

Step by step solution

01

Given

${鈭�}_{0}^{鈭�} x \ln x d x$

02

Calculate integral

Using by parts method of integration.

$鈭� u d v = u v - 鈭� v d u$

Here, $u = \ln x, d v = x$

$d u = \frac{1}{x}, v = \frac{x^{2}}{2}$

Substitute the above values,

${鈭�}_{0}^{鈭�} x \ln x d x = {[\frac{1}{2} x^{2} \ln x - 鈭� \frac{x^{2}}{2} \frac{1}{x} d x]}_{0}^{鈭�} = {[\frac{1}{2} x^{2} \ln x - \frac{1}{2} \frac{x^{2}}{2}]}_{0}^{鈭�} = {[\frac{1}{2} x^{2} \ln x - \frac{x^{2}}{4}]}_{0}^{鈭�} = 鈭� (Undefined)$

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Most popular questions from this chapter

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

Find three integrals in Exercises 21鈥�70 that we can anti-differentiate immediately after algebraic simplification.

Consider the integral $鈭� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$ from the reading at the beginning of the section.

(a) Use the inverse trigonometric substitution $u = \sin^{鈭� 1} 鈦� x$ to solve this integral.

(b) Use the trigonometric substitution $x = \sin 鈦� u$ to solve the integral.

(c) Compare and contrast the two methods used in parts (a) and (b).

Solve the integral

$鈭� \frac{1}{\sqrt{x^{2} + 9}} d x$

Solve the integral: $鈭� \frac{x}{e^{x}} d x$

See all solutions

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