Problem 16
Question
In Problems 1-16, evaluate each indefinite integral by making the given substitution. $$ \int \frac{x+1}{5-x} d x, \text { with } u=5-x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( -6 \ln |5-x| + 5 - x + C \).
1Step 1: Perform the Substitution
We start by substituting the given variable, set \( u = 5 - x \). Then differentiate to find \( dx \) in terms of \( du \). Since \( u = 5 - x \), it follows that \( -du = dx \).
2Step 2: Rewrite the Integral
Substitute \( u \) and \( dx \) into the integral. \( x = 5 - u \) from \( u = 5 - x \), so replace \( x \) with \( 5 - u \). The integral becomes: \[ \int \frac{(5-u)+1}{u} (-du) = \int \frac{6-u}{u} (-du) = -\int \left(\frac{6}{u} - 1\right) du. \]
3Step 3: Separate the Terms
Rewrite the integral by separating the terms in the fraction: \[ -\int \frac{6}{u} \, du + \int 1 \, du. \]
4Step 4: Integrate Each Term
Integrate each term separately. The integral \( \int \frac{6}{u} \, du = 6 \ln |u| \) and the integral \( \int 1 \, du = u \). Therefore, \[ - \left( 6 \ln |u| - u \right) = -6 \ln |u| + u. \]
5Step 5: Substitute Back in for x
Replace \( u \) with \( 5-x \) to convert back to terms of \( x \): \[ -6 \ln |5-x| + (5-x) + C, \] where \( C \) is the constant of integration.
Key Concepts
u-substitutiondifferentiationintegrationintegration constant
u-substitution
U-substitution is a very handy tool used in integration to simplify the function you need to integrate. In essence, it's about finding a new variable, commonly denoted as "u," to replace a more cumbersome expression in the integral.
The goal is to transform your integral into a form that's easier to work with. In our exercise, we started by letting \( u = 5 - x \). Once you define "u," you'll need to differentiate it with respect to \( x \) to express \( dx \) in terms of \( du \). For example, from \( u = 5 - x \), differentiating gives \( du = -dx \) or \( dx = -du \). This helps you rewrite the entire integral using "u."
The goal is to transform your integral into a form that's easier to work with. In our exercise, we started by letting \( u = 5 - x \). Once you define "u," you'll need to differentiate it with respect to \( x \) to express \( dx \) in terms of \( du \). For example, from \( u = 5 - x \), differentiating gives \( du = -dx \) or \( dx = -du \). This helps you rewrite the entire integral using "u."
- First, identify the part of the integral to substitute.
- Set \( u \) as that expression.
- Find \( du \) in terms of \( dx \) or vice versa.
- Transform the integral into terms of \( u \).
differentiation
Differentiation is intimately tied to u-substitution because it's used to find \( du \) from \( u \). Differentiation describes how a function changes as its input changes. It gives us rates at which function values are changing.
In our integral, after setting \( u = 5 - x \), we differentiate to find \( du \). The differentiation of \( u = 5 - x \) gives us \( -1 \) since the derivative of \( 5 \) is \( 0 \) and the derivative of \( -x \) is \( -1 \). Thus, we have \( du = -dx \), or rearranged, \( dx = -du \).
In our integral, after setting \( u = 5 - x \), we differentiate to find \( du \). The differentiation of \( u = 5 - x \) gives us \( -1 \) since the derivative of \( 5 \) is \( 0 \) and the derivative of \( -x \) is \( -1 \). Thus, we have \( du = -dx \), or rearranged, \( dx = -du \).
- Differentiate the expression for \( u \) with respect to \( x \).
- Use the derivative to express \( dx \) in terms of \( du \).
integration
Integration is the process of finding the original function from its derivative. It's essentially the reverse of differentiation. Once you've performed u-substitution and differentiation, integration becomes easier.
In our example, we've transformed the integral into two parts: \(-\int \frac{6}{u} \, du + \int 1 \, du\). Each part can be integrated separately:
Substitution and separation of terms simplify complex integrals into manageable calculations.
In our example, we've transformed the integral into two parts: \(-\int \frac{6}{u} \, du + \int 1 \, du\). Each part can be integrated separately:
- The integral \( \int \frac{6}{u} \, du \) results in \( 6 \ln |u| \).
- \( \int 1 \, du \) simply equals \( u \).
Substitution and separation of terms simplify complex integrals into manageable calculations.
integration constant
Every indefinite integral includes an arbitrary integration constant, commonly denoted as \( C \). Without any boundary conditions, it's impossible to determine the specific value of \( C \).
In our problem, after integrating and getting \( -6 \ln |u| + u \), you must add \( C \) because indefinite integrals represent families of functions, all differing only by a constant amount. This unknown constant allows for potential initial conditions or specific real-world values to update or specify the solution more precisely.
In our problem, after integrating and getting \( -6 \ln |u| + u \), you must add \( C \) because indefinite integrals represent families of functions, all differing only by a constant amount. This unknown constant allows for potential initial conditions or specific real-world values to update or specify the solution more precisely.
- Always include \( C \) when writing the solution to an indefinite integral.
- Remember, \( C \) reflects the freedom of having infinite functions that share the same derivative.
Other exercises in this chapter
Problem 16
In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x \csc ^{2} x d x $$
View solution Problem 16
Use long division to write \(f(x)\) as a sum of a polynomial and a proper rational function. $$ f(x)=\frac{x^{2}+1}{3 x+1} $$
View solution Problem 17
(a) Find the Taylor polynomial of degree 3 about \(x=0\) for \(f(x)=\sin x\). (b) Use your result in (a) to explain why $$ \lim _{x \rightarrow 0} \frac{\sin x}
View solution Problem 17
Use a spreadsheet to approximate each of the following integrals using the midpoint rule with each of the specified values of \(n .\) \(\int_{0}^{1} x^{2} d x\)
View solution