/*! 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 49 Find the indefinite integral. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the indefinite integral. $$ \int \frac{1}{2 x+3} d x $$

Short Answer

Expert verified
The indefinite integral of the given function is: \(\int \frac{1}{2 x+3} d x = \frac{1}{2} \cdot ln |2x + 3| + C\)

Step by step solution

01

Identify u

In this problem, we can see that the denominator is a linear function of x, 2x + 3. We will set this as our u: \(u = 2x + 3\) Step 2: Find du
02

Find du

Next, we need to find the derivative of u with respect to x, denoted as du/dx. \(\frac{d u}{d x} = \frac{d(2 x+3)}{d x} = 2\) Now find du: \(d u = 2 d x\) Step 3: Substitute u and du
03

Substitute u and du

Replace 2x + 3 with u and dx with du/2: \(\int \frac{1}{u} \cdot \frac{d u}{2}\) Step 4: Solve the integral
04

Solve the integral

Simplify and integrate: \(\frac{1}{2} \int \frac{1}{u} d u\) The integral of \(\frac{1}{u}\) with respect to u is \(ln |u|\). Therefore, we have: \(\frac{1}{2} \cdot ln |u| + C\), where C is the constant of integration. Step 5: Replace u with the original expression
05

Replace u with the original expression

Finally, substitute back the original expression of u: \(\frac{1}{2} \cdot ln |2x + 3| + C\) The indefinite integral of the given function is: \(\int \frac{1}{2 x+3} d x = \frac{1}{2} \cdot ln |2x + 3| + C\)

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Key Concepts

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

u-substitution
The u-substitution method is a popular technique for finding indefinite integrals because it simplifies the integrand, making integration easier. In this method, we substitute a part of the integral with a new variable "u". This substitution typically involves the inner function of a composite function.For the given integral \( \int \frac{1}{2x+3} \, dx \), we identified \( u = 2x + 3 \). Next, we find the derivative of \( u \) with respect to \( x \), which is \( du = 2 \, dx \). This allows us to rewrite the integral in a new, simpler form. The goal is to transform the integral into a basic form so we can solve it easily. After finishing the integration process, we reverse the substitution by plugging back the expression in terms of the original variable \( x \). This method is particularly effective for rational functions and helps in evaluating otherwise complex integrals.
integration techniques
Integration techniques encompass a variety of methods used to solve integrals, each suited for different types of functions. u-Substitution, as discussed, is one technique used for functions that resemble derivatives of known functions. Other common techniques include:
  • Integration by Parts, used for the product of functions.
  • Trigonometric Substitution, applicable for integrals involving root expressions.
  • Partial Fraction Decomposition, used for simplifying rational functions into simpler fractions.
For our exercise, we used u-substitution to handle the linear denominator, transforming the problem efficiently into a basic integral form. Recognizing when and how to use these techniques is essential, as some integrals might seem complex but can be simplified using the right approach. These methods are powerful tools in calculus, each tailored to tackle specific integral types.
logarithmic integration
Logarithmic integration is a technique that often arises in integrals of the form \( \int \frac{1}{u} \, du \). This type of integral results in the natural logarithm function. In our exercise, after applying u-substitution, the integral reduced to this form, leading to:\[ \int \frac{1}{u} \, du = \ln |u| + C \]The presence of the absolute value symbol \( |u| \) ensures the logarithm remains valid for negative inputs, maintaining it as a real number. This technique finds wide application in problems where the derivative of a logarithm is repositioned within the integrand. Logarithmic integration simplifies the complex relationship of exponential and reciprocal functions into a manageable form, making integrations involving simple fractions straightforward.
constants of integration
When dealing with indefinite integrals, we follow up our solution with a constant of integration, denoted \( C \). This constant accounts for any constants that originally existed in the function before its derivative was taken. Since derivatives of constant terms are zero, any number could be added to the antiderivative, thus requiring the constant \( + C \) in our final answer. This renders our solution\[ \frac{1}{2} \ln |2x + 3| + C \]adjustable for various conditions. It's imperative to include this constant, as it signifies the infinite family of functions derived from an antiderivative. Understanding and properly applying the constant of integration ensures precise evaluations in indefinite integrals, aligning them with initial conditions in applications like physics and engineering.

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Most popular questions from this chapter

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