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Chapter 7: Problem 62

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{1+\sqrt{x}}{1-\sqrt{x}} d x $$

Short Answer

Expert verified
Use substitution \(u=\sqrt{x}\), simplify, integrate by parts if needed, and verify with CAS.

Step by step solution

01

Identify the Substitution

We will use the substitution method. Notice that the expression contains \(\sqrt{x}\), which suggests a substitution to simplify the square root. Let's choose \(u = \sqrt{x}\), which implies \(u = x^{1/2}\). Thus, \(du = \frac{1}{2\sqrt{x}}\, dx\) or \(dx = 2u\, du\).
02

Rewrite the Integral

Using the substitution \(u = \sqrt{x}\), rewrite the integral. The integral becomes:\[ \int \frac{1 + u}{1 - u} \cdot 2u \, du \]. Distribute the \(2u\) in the numerator, leading to:\[ \int \frac{2u(1 + u)}{1-u} \, du \].
03

Simplifying the Expression

Simplify the expression inside the integral:\[ \int \frac{2u + 2u^2}{1-u} \, du = \int \left( \frac{2u}{1-u} + \frac{2u^2}{1-u} \right) \, du \]. This separates into two integrals:\[ 2\int \frac{u}{1-u} \, du + 2\int \frac{u^2}{1-u} \, du \].
04

Solve Each Integral

Now we will solve each integral separately. For the first integral \(2\int \frac{u}{1-u} \, du\):Use the substitution \(v = 1-u\), then \(dv = -du\),\(u = 1-v\). The integral becomes:\[ -2\int \frac{1-v}{v} \, dv \].This can be simplified and split:\[ -2\int \left(\frac{1}{v}-1\right) \, dv = -2(\ln |v| - v) + C_1 = -2(\ln |1-u| - (1-u)) + C_1 \].For the second integral \(2\int \frac{u^2}{1-u} \, du\),Direct integration isn't straightforward, so we apply integration by parts or rewrite if necessary based on further substitutions.Note: This step can be complex and might require adjustment based on advanced calculus techniques.
05

Combine Results and Back-Substitute

Add the results of the two integrals from the previous step and back-substitute using \(u = \sqrt{x}\).\( -2(\ln |1-\sqrt{x}| - (1-\sqrt{x})) + C_1\).Ensure that combined and simplified results reflect substitution back to original variable \(x\).
06

Confirm using CAS

Utilize a Computer Algebra System (CAS) to calculate:\[ \int \frac{1+\sqrt{x}}{1-\sqrt{x}} \, dx \].Compare it with our solution to confirm both results match.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Substitution
Integration by substitution is a powerful technique used to simplify difficult integrals. By substituting a part of the integral with a new variable, you can often convert the integral into a more manageable form.

In our exercise, the expression includes a square root, \( \sqrt{x} \). This indicates that substituting \( u = \sqrt{x} \) could simplify the problem. This choice leads us to a derivative of \( x \) with respect to \( u \): \( dx = 2u \, du \).
  • Transforming the integral using \( u = \sqrt{x} \), the parts inside the integral get expressed in terms of \( u \) instead of \( x \).
  • This substitution can simplify operations, such as simplifying rational expressions or reducing powers.
  • By changing the variable, it can also make the integration process easier to visualize and execute.
Understanding when and what to substitute is crucial in integral calculus, as the right substitution can transform a complex integration into a straightforward one.
Square Root Simplification
Square root simplification plays a critical role when dealing with expressions involving radicals, like \( \sqrt{x} \). By substituting \( u = \sqrt{x} \), the troublesome radical becomes a simple linear variable \( u \), significantly simplifying the integral.

The idea is to eliminate the square root so as to make the expression easier to handle:
  • Substitution like \( u = \sqrt{x} \) changes square roots to linear expressions which are easier to integrate.
  • In this exercise, \( dx \) gets expressed in terms of \( du \), making the integration operation more seamless.
This form of simplification is particularly effective in integrals where radicals are in the numerator or denominator of a fraction, as it helps manage the variable's complexity much better. Recognizing when to use such substitution for radical expressions is key to simplifying integrals.
Integration Techniques
Integration techniques often require multiple strategies depending on the form of the integral. Combining methods like substitution, simplification, and recognizing standard forms will help solve various integrals efficiently.

In the given problem, breaking down the problem step-by-step is vital.
  • After substitution, distributing and separating the integral into simpler parts can make tackling each section more straightforward.
  • For example, in our case, the initial integral transformed into two separate integrals which could be approached individually.
  • If necessary, these parts could further use techniques like integration by parts or advanced algebraic manipulations to simplify.
Every integral presents unique challenges, and learning these techniques allows students to apply the right method at the right time. Mastering integration is about developing these tactical approaches to deal with different kinds of problems.

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Most popular questions from this chapter

Transform the given improper integral into a proper integral by making the stated \(u\) -substitution; then approximate the proper integral by Simpson's rule with \(n=10\) subdivisions. Round your answer to three decimal places. $$ \int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x ; u=\sqrt{x} $$

(a) Make \(u\) -substitution (5) to convert the integrand to a rational function of \(u\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral (no substitution), and then confirm that the result is equivalent to that in part (a). $$ \int \frac{d x}{\sin x+\tan x} $$

(a) Make an appropriate \(u\) -substitution of the form \(u=x^{1 / n}\) or \(u=(x+a)^{1 / n}\), and then evaluate the integral. (b) If you have a CAS, use it to evaluate the integral, and then confirm that the result is equivalent to the one that you found in part (a). $$ \int \frac{d x}{x^{1 / 2}-x^{1 / 3}} $$

It is sometimes possible to convert an improper integral into a "proper" integral having the same value by making an appropriate substitution. Evaluate the following integral by making the indicated substitution, and investigate what happens if you evaluate the integral directly using a CAS. $$ \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x ; u=\sqrt{1-x} $$

Determine whether the statement is true or false. Explain your answer. The Simpson's rule approximation \(S_{50}\) for \(\int_{a}^{b} f(x) d x\) is a weighted average of the approximations \(M_{50}\) and \(T_{50}\), where \(M_{50}\) is given twice the weight of \(T_{50}\) in the average.
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