/*! 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 43 Antidifferentiate using the tabl... [FREE SOLUTION] | 91Ó°ÊÓ

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Antidifferentiate using the table of integrals. You may need to transform the integrand first. $$\int \frac{1}{x^{2}+4 x+3} d x$$

Short Answer

Expert verified
\( \frac{1}{2} \ln \left( \frac{|x+1|}{|x+3|} \right) + C \)

Step by step solution

01

Simplify the Integrand

The integrand is \( \frac{1}{x^{2}+4x+3} \). Notice that the denominator can be factored as \( (x+1)(x+3) \). This suggests that partial fraction decomposition might be a suitable technique.
02

Set Up Partial Fraction Decomposition

Express \( \frac{1}{x^{2}+4x+3} \) as the sum of partial fractions. We assume it can be written as \( \frac{A}{x+1} + \frac{B}{x+3} \).
03

Solve for Constants A and B

To solve for \( A \) and \( B \), clear the fractions by multiplying through by \( (x+1)(x+3) \):\[ 1 = A(x+3) + B(x+1) \]Expand and collect like terms:\[ 1 = (A + B)x + (3A + B) \]There are no \( x \) terms on the left, so \( A + B = 0 \). Also, \( 3A + B = 1 \). Solving these equations, we find \( A = \frac{1}{2} \), \( B = -\frac{1}{2} \).
04

Rewrite the Integral Using Partial Fractions

Substitute the values of \( A \) and \( B \) back into the decomposition:\[ \int \frac{1}{x^{2}+4x+3} \, dx = \int \left( \frac{1}{2(x+1)} - \frac{1}{2(x+3)} \right) \, dx \]
05

Integrate Each Term Separately

Integrate each term separately. Using the integral formula for \( \int \frac{1}{x} \, dx = \ln|x| + C \), find:\[ \int \left( \frac{1}{2(x+1)} - \frac{1}{2(x+3)} \right) \, dx = \frac{1}{2} \ln|x+1| - \frac{1}{2} \ln|x+3| + C \]
06

Simplify the Final Result

Combine the logarithms using the properties of logarithms:\[ \frac{1}{2} \ln|x+1| - \frac{1}{2} \ln|x+3| = \frac{1}{2} \ln \left( \frac{|x+1|}{|x+3|} \right) \]Thus, the antiderivative is:\[ \frac{1}{2} \ln \left( \frac{|x+1|}{|x+3|} \right) + C \]
07

Verify the Result

To verify, differentiate the result: \( \frac{d}{dx} \left[ \frac{1}{2} \ln \left( \frac{|x+1|}{|x+3|} \right) \right] \) yields \( \frac{1}{x^{2}+4x+3} \), confirming our antiderivative is correct.

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Key Concepts

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

Partial Fraction Decomposition
When faced with the integration of a rational function where the integral isn't immediately recognizable, partial fraction decomposition can simplify the problem. This method involves breaking down a complex fraction into simpler components, making it easier to integrate. For instance, if we have a rational expression like \( \frac{1}{x^{2}+4x+3} \), we can decompose it into simpler fractions.
The first step is to factor the denominator, \( x^2 + 4x + 3 \), into two linear factors: \( (x+1) \) and \( (x+3) \). This decomposition helps rewrite the expression as \( \frac{A}{x+1} + \frac{B}{x+3} \), where \( A \) and \( B \) are constants that we will solve for.
To find these constants, we set up an equation by multiplying through by the denominator \( (x+1)(x+3) \), clearing the fractions. Solving the resulting equations gives us values for \( A \) and \( B \). This method is an essential integration technique, especially for rational functions, where instead of one difficult integral, we solve two or more simpler integrals.
Integration Techniques
Once the expression is broken down using partial fraction decomposition, we apply suitable integration techniques to each term. Let's consider the terms \( \frac{A}{x+1} \) and \( \frac{B}{x+3} \) that emerged from our decomposition.
These terms are integrated using standard formulas. Specifically, the integral of \( \frac{1}{x} \) is well-known as \( \ln|x| + C \). Thus, the integral of \( \frac{1}{2(x+1)} \) becomes \( \frac{1}{2} \ln|x+1| \) and the integral of \( \frac{-1}{2(x+3)} \) is \( -\frac{1}{2} \ln|x+3| \).
In general, integration techniques vary widely depending on the function at hand. Familiarity with such techniques aids significantly in solving calculus problems. While partial fraction decomposition is a powerful tool, understanding when to use each technique is crucial for learning integration.
Integral of Rational Functions
The integration of rational functions often requires breaking down the function into simpler parts before proceeding. Rational functions are quotients of polynomials, and when dividing yields complex terms, the process of integration becomes tricky.
  • Partial Fraction Decomposition: This is often the first step to simplify a rational function, involving splitting it into simpler fractions.
  • Integration by Parts: Sometimes used alongside partial fractions when encountering more complex polynomials.
  • Trigonometric Substitution: In some cases, these methods are applied when the simplified terms involve quadratic polynomials.
Understanding these methods and knowing when to apply them is vital when dealing with integrals of rational functions. By thorough practice and application of these techniques, one can gain proficiency in integrating rational expressions and other complex integrals encountered in calculus.

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Most popular questions from this chapter

Solve without using integral tables. Calculate the integral \(\int \frac{1}{(x-a)(x-b)} d x\) for (a) \(a \neq b\) (b) \(a=b\)

(a) Find \(\int \sin \theta \cos \theta d \theta\) (b) You probably solved part (a) by making the substitution \(w=\sin \theta\) or \(w=\cos \theta .\) (If not, go back and do it that way.) Now find \(\int \sin \theta \cos \theta d \theta\) by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities $$ \begin{aligned}\sin (2 \theta) &=2 \sin \theta \cos \theta \\\\\cos (2 \theta) &=\cos ^{2} \theta-\sin ^{2} \theta \end{aligned}$$ Find \(\int \sin \theta \cos \theta d \theta\) using one of these identities and then the substitution \(w=2 \theta\) (d) You should now have three different expressions for the indetinite integral \(\int \sin \theta \cos \theta d \theta .\) Are they really different? Are they all correct? Explain.

Decide whether the statements are true or false. Give an explanation for your answer. \(\int 1 /\left(x^{2}+4 x-5\right) d x\) involves an arctangent.

With \(t\) in years since April 20 th, \(2015,\) the population. \(P .\) of the world in billions \(^{2}\) can be modeled by \(P=7.17 e^{0.01064 t}\) (a) What does this model predict for the world population on April 20 th, \(2020 ?\) In \(2025 ?\) (b) Use the Fundamental Theorem to predict the average population of the world between April 2015 and April 2025.

Explain why the two antiderivatives are really, despite their apparent dissimilarity, different expressions of the same problem. You do not need to evaluate the integrals. $$\int e^{\sin x} \cos x d x \quad \text { and } \quad \int \frac{e^{\arcsin x}}{\sqrt{1-x^{2}}} d x$$

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