/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Determine the following indefini... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 53

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{1}{x \sqrt{x^{2}-25}} d x$$

Short Answer

Expert verified
Question: Find the indefinite integral of the function \(f(x) = \frac{1}{x\sqrt{x^2-25}}\). Answer: The indefinite integral of the function is \(F(x) = \sqrt{x^2 - 25} + C\), where C is the constant of integration.

Step by step solution

01

Identify Substitution Candidate

Observe the expression under the integral and identify the candidate for substitution. In this case, the best option is to set \(u = x^2 - 25\), so that \(d u = 2x \, d x\).
02

Rewrite the Integral with Substitution

Using \(u = x^2 - 25\), we have \(d u = 2x \, d x\). Now, we substitute into the integral: $$ \int \frac{1}{x\sqrt{x^2-25}}\, dx = \frac{1}{2} \int \frac{1}{\sqrt{u}}\, du $$
03

Integrate with Respect to u

Now, we calculate the integral of the substituted function with respect to \(u\): $$ \frac{1}{2} \int \frac{1}{\sqrt{u}}\,du = \frac{1}{2} \int u^{-\frac{1}{2}}\,du $$ Applying the basic integration rule, we get: $$ \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = u^{\frac{1}{2}} + C . $$
04

Rewrite Integral in Terms of x

Now that we have found the integral with respect to \(u\), we substitute the original expression for \(u\) to rewrite the integral in terms of \(x\): $$ u^{\frac{1}{2}} + C = (x^2 - 25)^{\frac{1}{2}} + C = \sqrt{x^2 - 25} + C $$ So, the indefinite integral is: $$ \int \frac{1}{x\sqrt{x^2-25}}\, dx = \sqrt{x^2 - 25} + C $$
05

Check Work by Differentiation

Now we need to check our work by differentiating the result and see if we get back the original function under the integral: $$ \frac{d}{d x}\left(\sqrt{x^2 - 25} + C\right) $$ Using the chain rule, we get: $$ \frac{1}{2}(x^2 - 25)^{-\frac{1}{2}} \cdot 2x = \frac{x}{\sqrt{x^2 - 25}} $$ This is equal to the original expression under the integral, confirming that our indefinite integral is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the function \(F\) that satisfies the following differential equations and initial conditions. $$F^{\prime \prime \prime}(x)=672 x^{5}+24 x, F^{\prime \prime}(0)=0, F^{\prime}(0)=2, F(0)=1$$

Determine whether the following statements are true and give an explanation or counterexample. a. If \(f^{\prime}(x)>0\) and \(f^{\prime \prime}(x)<0\) on an interval, then \(f\) is increasing at a decreasing rate on the interval. b. If \(f^{\prime}(c)>0\) and \(f^{\prime \prime}(c)=0,\) then \(f\) has a local maximum at \(c\) c. Two functions that differ by an additive constant both increase and decrease on the same intervals. d. If \(f\) and \(g\) increase on an interval, then the product \(f g\) also increases on that interval. e. There exists a function \(f\) that is continuous on \((-\infty, \infty)\) with exactly three critical points, all of which correspond to local maxima.

Let \(a\) and \(b\) be positive real numbers. Evaluate \(\lim _{x \rightarrow \infty}(a x-\sqrt{a^{2} x^{2}-b x})\) in terms of \(a\) and \(b\).

The functions \(f(x)=a x^{2},\) where \(a>0\) are concave up for all \(x\). Graph these functions for \(a=1,5,\) and 10, and discuss how the concavity varies with \(a\). How does \(a\) change the appearance of the graph?

A mass oscillates up and down on the end of a spring. Find its position \(s\) relative to the equilibrium position if its acceleration is \(a(t)=\sin (\pi t),\) and its initial velocity and position are \(v(0)=3\) and \(s(0)=0,\) respectively.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.