/*! 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 47 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals in Exercises \(39-50\) $$\begin{array}{l}{\int \frac{\sqrt{x+1}}{x} d x} \\ {\text { (Hint. Let } x+1=u^{2} . )}\end{array}$$

Short Answer

Expert verified
The integral evaluates to \( 2\sqrt{x+1} + \ln|\sqrt{x+1}-1| - \ln|\sqrt{x+1}+1| + C \).

Step by step solution

01

Substitute Variable

Use the substitution provided in the hint. Let \( x + 1 = u^2 \). Then, differentiate both sides: \( dx = 2u \, du \). Notice that as \( x \) changes, \( x+1 \) becomes \( u^2 \), altering the square root component in the integral.
02

Change of Variable

Substitute \( x + 1 = u^2 \) and \( dx = 2u \, du \) into the integral:\[ \int \frac{\sqrt{x+1}}{x} \, dx = \int \frac{u}{u^2 - 1} \, 2u \, du \]This simplifies to:\[ 2 \int \frac{u^2}{u^2 - 1} \, du \]
03

Simplify the Integrand

Separate the integrand into simpler parts:\[ \frac{u^2}{u^2 - 1} = 1 + \frac{1}{u^2 - 1} \]Thus, the integral becomes:\[ 2 \int \left( 1 + \frac{1}{u^2 - 1} \right) \, du \]
04

Integrate Each Term

Integrate the separated terms individually:1. \( \int 1 \, du = u \)2. Use partial fraction decomposition on \( \int \frac{1}{u^2 - 1} \, du \). Write: \[ \frac{1}{u^2 - 1} = \frac{1}{2} \left( \frac{1}{u-1} - \frac{1}{u+1} \right) \]The integral becomes:\[ \int \frac{1}{u^2 - 1} \, du = \frac{1}{2} \ln|u-1| - \frac{1}{2} \ln|u+1| \]
05

Combine Integrals

Combine the results from integrating both terms:\[ 2 \left( u + \frac{1}{2} \ln|u-1| - \frac{1}{2} \ln|u+1| \right) = 2u + \ln|u-1| - \ln|u+1| \]
06

Substitute Back to Original Variable

Return to the original variable using \( u = \sqrt{x+1} \). Substitute back:\[ 2\sqrt{x+1} + \ln|\sqrt{x+1}-1| - \ln|\sqrt{x+1}+1| + C \]Here, \( C \) is the constant of integration.

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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 in calculus that simplifies the process of integrating complex functions. It's akin to reversing the chain rule for derivatives. The core idea is to transform a complicated integral into a simpler one using a suitable substitution.
For instance, in our problem, we substitute the expression inside the square root, which is often a good candidate for making integration easier. Let's say we have a term like \( \sqrt{x+1} \). Here, we set \( x + 1 = u^2 \), turning the problem into a form that could be easier to handle. Once you perform this substitution, the expression \( \sqrt{x+1} \) simply becomes \( u \), and the differential \( dx \) becomes \( 2u \, du \).
  • Start by identifying the part of the integrand that can be substituted to make integration simpler.
  • Replace all \( x \)-related parts with expressions related to \( u \).
  • Don't forget to also change the limits of integration if you're working with definite integrals.
By using substitution, complex expressions reduce to simpler integrals, enabling us to compute them more easily.
Partial Fraction Decomposition
Partial fraction decomposition is another essential tool in integral calculus, especially when dealing with rational functions. A rational function is one polynomial divided by another. The goal of decomposition is to break down complex fractions into easier, simpler pieces that are more straightforward to integrate.
When faced with the function \( \frac{1}{u^2-1} \), as in our example, we recognize an opportunity to apply this method. We express \( \frac{1}{u^2-1} \) as a sum of simpler fractions. Notice that \( u^2 - 1 \) can be factored as \((u-1)(u+1)\), leading to:
\[ \frac{1}{u^2-1} = \frac{1}{2} \left( \frac{1}{u-1} - \frac{1}{u+1} \right) \]
  • First, factor the denominator if possible.
  • Express the original rational function as a sum of partial fractions.
  • Each component fraction should have a simpler denominator than the original.
This decomposition allows the integral to be separated into simpler parts, which can each be integrated easily, like logarithmic integrals in this particular case.
Change of Variable
The change of variable technique closely ties in with substitution. It helps us rewrite the integral in a completely different form, both simplifying it and making it feasible to solve. This transformation is especially useful when a direct integration looks difficult or is not possible.
In the problem, initially our integral, \( \int \frac{\sqrt{x+1}}{x} \, dx \), is challenging because of the roots and the division by \( x \). By substituting \( x+1 = u^2 \), we carry out the change of variable which greatly simplifies our original integral into \( \int \frac{u}{u^2 - 1} \, 2u \, du \).
  • Identify substitution opportunities that will transform an uncomfortable integral.
  • Replace not only the function but also differentials, like \( dx \) to \( 2u \, du \), consistent with the substitution.
  • Simplifying such integrals lays the foundation for easy solving of the rest of the problem.
In essence, the change of variable acts like a mathematical tool that reshapes the integrand into a form that is much easier to handle using standard calculus techniques.

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Most popular questions from this chapter

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