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Chapter 7: Problem 40

Make a substitution to express the integrand as a rational function and then evaluate the integral. \( \displaystyle \int \frac{dx}{2 \sqrt{x + 3} + x} \)

Short Answer

Expert verified
Substitute \( u = \sqrt{x+3} \) and transform the integral to evaluate it.

Step by step solution

01

Identify a Suitable Substitution

Let's start by recognizing that the expression under the square root can be reorganized using substitution. We'll set a new variable, say, \( u = \sqrt{x + 3} \). From this, we derive \( x = u^2 - 3 \).
02

Differentiate the Substitution

To change variables, differentiate the substitution. If \( u = \sqrt{x+3} \), differentiating gives us \( du = \frac{1}{2\sqrt{x+3}} dx \) or equivalently, \( dx = 2u \, du \).
03

Express the Integrand in Terms of u

Substitute \( x = u^2 - 3 \) and \( dx = 2u \, du \) back into the original integral. The integrand becomes:\[\int \frac{2u \, du}{2u + (u^2 - 3)}\]This simplifies to:\[\int \frac{2u \, du}{u^2 + 2u - 3}\]
04

Complete the Square for the Denominator

Notice that the expression \( u^2 + 2u - 3 \) can be completed to form a perfect square. It becomes:\[u^2 + 2u - 3 = (u + 1)^2 - 4\]
05

Perform Partial Fraction Decomposition

We can express the fraction using partial fractions. The denominator \((u+1)^2 - 4\) can be factored as \((u - 1)(u + 3)\). Write:\[\frac{2u}{(u - 1)(u + 3)} = \frac{A}{u - 1} + \frac{B}{u + 3}\]Solve for \(A\) and \(B\) by multiplying through and equating coefficients.
06

Integrate the Rational Function

After finding \(A\) and \(B\), integrate each term separately:\[\int \frac{A}{u - 1} \, du + \int \frac{B}{u + 3} \, du\]These yield logarithmic solutions. Combine them to find the whole solution in terms of \(u\).
07

Substitute Back to x

Substitute \( u = \sqrt{x + 3} \) back into the integrated result to express the integral in terms of \(x\). Simplify the result if possible.

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Key Concepts

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

Substitution Method
The substitution method is a fundamental technique used in calculus to simplify and evaluate complex integrals. It involves substituting a part of the integrand with a new variable, making the integral easier to solve. In our exercise, we needed to make the integral more manageable by choosing an appropriate substitution.
  • First, identify a part of the integrand that can be replaced. Here, we chose to substitute the expression under the square root.
  • The substitution made was: \( u = \sqrt{x + 3} \), transforming the integral significantly.
  • By differentiating the substitution, we found \( dx = 2u \, du \), allowing us to replace \( dx \) in terms of \( du \).
Utilizing the substitution method effectively can convert a challenging integral into a form that is much more straightforward to evaluate.
Rational Functions
A rational function is any function that can be expressed as the ratio of two polynomials. Understanding rational functions is key to solving the given integral after substitution converted it into a rational expression.
  • After substituting, our integral becomes: \( \int \frac{2u \, du}{u^2 + 2u - 3} \)
  • This expression is rational because it's a polynomial in the numerator divided by another polynomial in the denominator.
Working with rational functions in calculus often involves techniques like polynomial division or partial fraction decomposition to simplify the process of integration.
Partial Fraction Decomposition
Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to integrate. In this exercise, it was applied to the integrand after substitution.
  • The denominator, \( u^2 + 2u - 3 \), was rewritten through factoring as \((u - 1)(u + 3)\). This allowed for decomposition.
  • We expressed \( \frac{2u}{(u - 1)(u + 3)} \) as \( \frac{A}{u - 1} + \frac{B}{u + 3} \) by solving for \( A \) and \( B \).
  • Once the rational function is decomposed, each simpler fraction can be integrated separately, simplifying the overall integral computation.
This technique simplifies complex fractions, making integration more straightforward and manageable.
Definite Integrals
Definite integrals are a key concept in calculus that give the signed area under a curve in an interval and can determine accumulated quantities. While this exercise focuses on indefinite integration, understanding definite integrals aids in seeing the broader application of these techniques.
  • Definite integrals provide a numerical result, associated with the limits of integration \( a \) and \( b \), using the Fundamental Theorem of Calculus.
  • If integration limits were given, evaluating the result from \( a \) to \( b \) would replace \( x \).
  • This topic is crucial for problems involving accumulation, such as area, volume, and in real-world applications.
Mastering the transition from indefinite to definite integrals is vital for solving practical calculus problems that involve specific numeric results.

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Most popular questions from this chapter

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \( \displaystyle \int_0^4 \frac{dx}{x^2 - x - 2} \)

Use (a) the Midpoint Rule and (b) Simpson's Rule to approximate the given integral with the specified value of \( n \). (Round your answers to six decimal places.) Compare your results to the actual value to determine the error in each approximation. \( \displaystyle \int_0^2 \frac{x}{1 + x^2}\ dx \) , \( n = 10 \)

Explain why each of the following integrals is improper. (a) \( \displaystyle \int_1^2 \frac{x}{x - 1}\ dx \) (b) \( \displaystyle \int_0^\infty \frac{1}{1 + x^3}\ dx \) (c) \( \displaystyle \int_{-\infty}^\infty x^2 e^{-x^2}\ dx \) (d) \( \displaystyle \int_0^{\frac{\pi}{4}} \cot x\ dx \)

Use a computer algebra system to evaluate the integral. Compare the answer with the result of using tables. If the answers are not the same, show that they are equivalent. \( \displaystyle \int \frac{1}{\sqrt{1 + \sqrt[3]{x}}}\ dx \)

Sketch the region and find its area (if the area is finite). \( S = \\{ (x, y) \mid x \ge 1, 0 \le y \le \frac{1}{(x^3 + x)} \\} \)
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