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Chapter 5: Q. 56 (page 496)

$\int_{0}^{1} \frac{1}{x {(\ln x)}^{2}} d x$

Short Answer

Expert verified

Undefined

The limit does not converge.

Step by step solution

01

Given

$\int_{0}^{1} \frac{1}{x {(\ln x)}^{2}} d x$

02

Calculate integral

By using the substitution method,

Let,

$\ln x = t \frac{1}{x} d x = d t d x = x d t$

When,

$x = 0 \Rightarrow t = - \infty x = 1 \Rightarrow t = \ln (1) = 0$

$\int_{0}^{1} \frac{1}{x {(\ln x)}^{2}} d x = \int_{- \infty}^{0} \frac{1}{t^{2}} d t = {[\frac{t^{- 2 + 1}}{- 2 + 1}]}_{- \infty}^{0} = - {[\frac{1}{t}]}_{- \infty}^{0} = - (\frac{1}{0} + \frac{1}{\infty}) = - \infty (Undefined)$

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Most popular questions from this chapter

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.

$\int \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{x^{2} - 4} d x$ .

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{\sqrt{x^{2} - 4}} d x$ .

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving $\int \frac{1}{x^{2} + 4} d x .$

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving $\int {(x^{2} + 4)}^{- 5 / 2} d x$

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form $\sqrt{x^{2} - a^{2}}$ .

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with $x = a \sin u$ , we must consider the cases $x > a$ and $x < - a$ separately.

(h) True or False: When using trigonometric substitution with $x = a s e c u$ , we must consider the cases $x > a$ and $x < - a$ separately.

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.

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

Suppose you use polynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial $x^{2} - x + 3$ as the quotient above the top line, and the polynomial 3x − 1 at the bottom as the remainder. Then $p (x) =___a n d \frac{p (x)}{q (x)} =____$

Solve $\int \frac{x + 2}{{(x^{2} + 4 x)}^{\frac{3}{2}}} d x$ the following two ways:

(a) with the substitution $u = x^{2} + 4 x;$

(b) by completing the square and then applying the trigonometric substitution x + 2 = 2 sec u.

See all solutions

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