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Chapter 7: Problem 32

Find the integrals Check your answers by differentiation. $$\int \frac{e^{t}+1}{e^{t}+t} d t$$

Short Answer

Expert verified
The integral is \( \ln |e^t + t| + C \).

Step by step solution

01

Check the Integral for Simple Methods

Look at the integral \( \int \frac{e^t + 1}{e^t + t} \, dt \). First, determine if there are any straightforward substitutions or simplifications to apply. Notice that simple algebra doesn't appear to simplify it.
02

Consider Substitution

Attempt substitution by setting \( u = e^t + t \), then \( du = (e^t + 1) \, dt \). This makes the integral \( \int \frac{du}{u} \), which is simpler.
03

Integrate Using Substitution

Recognize that \( \int \frac{du}{u} = \ln |u| + C \). Replace \( u \text{ with } (e^t + t) \). Thus, \( \int \frac{e^t + 1}{e^t + t} \, dt = \ln |e^t + t| + C \).
04

Verify by Differentiating

Differentiate \( \ln |e^t + t| + C \) with respect to \( t \). The derivative is \( \frac{1}{e^t + t} \cdot (e^t + 1) \), which matches the original integrand \( \frac{e^t + 1}{e^t + t} \), confirming the integration.

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Key Concepts

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

Integration Techniques
Integration is a key concept in calculus that enables us to find the area under a curve and solve equations involving derivatives. When approaching an integral, there are several techniques you might use:
  • Substitution: This helps in simplifying the integral by changing variables, essentially transforming it into a more manageable form.
  • Integration by Parts: Useful when dealing with integrands that are products of functions. It is based on the reverse application of the product rule of differentiation.
  • Trigonometric Integrals: Applying trigonometric identities can simplify integrals that involve trigonometric functions.
  • Partial Fraction Decomposition: Useful for rational functions, where you express the function as a sum of simpler fractions.
Once you have an integral, start by looking for a technique that fits it best. In our problem, substitution is utilized because the form of the integrand suggests it could simplify nicely. Always consider verifying your result. Differentiation provides a great way to check your work, ensuring the original function is retrieved.
Differentiation
In calculus, differentiation is the process of finding the derivative of a function. A derivative represents how a function changes as its input changes. Here are some basics to remember:
  • The power rule works for monomials and is expressed as: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
  • The product rule states: for two functions \( u(x) \) and \( v(x) \), \( (uv)' = u'v + uv' \).
  • The chain rule is crucial when dealing with composite functions: if \( y = g(f(x)) \), then \( \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \).
In the context of verifying an integral result, differentiation plays a verification role. Solving our problem involves distinguishing every component and checking if the differentiated result equals the original integrand. This ensures that the integration was performed correctly.
Substitution Method
The substitution method, often referred to as "u-substitution," is a useful technique in integrating functions. Its goal is to make integration more straightforward by switching to a new variable that simplifies the integrand.
  • Identify a part of the integrand to replace with a new variable \( u \). This choice can simplify the expression greatly.
  • Compute \( du \), the differential of \( u \), in terms of \( dx \); this part helps in changing the variable of integration.
  • Rewrite the entire integral in terms of \( u \). This new integral should be easier to solve.
For the original exercise, setting \( u = e^t + t \) turned the complex integrand into a simple natural log function. After solving the transformed integral, substitute back the original variable to get the final result. This method is extremely effective when a function or its derivative already appears within the integral.

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Most popular questions from this chapter

Let \(f(t)\) be the rate of flow, in cubic meters per hour, of a flooding river at time \(t\) in hours. Give an integral for the total flow of the river (a) Over the 3 -day period \(0 \leq t \leq 72\) (b) In terms of time \(T\) in days over the same 3 -day period.

Explain what is wrong with the statement. The table can be used to evaluate \(\int \sin x / x \, d x\)

Find a substitution \(w\) and constants \(a, b, k\) so that the integral has the form \(\int_{a}^{b} k f(w) d w\). $$\int_{1}^{9} f(6 x \sqrt{x}) \sqrt{x} d x$$

Integrate: (a) \(\int \frac{1}{\sqrt{x}} d x\) (b) \(\int \frac{1}{\sqrt{x+1}} d x\) (c) \(\int \frac{1}{\sqrt{x}+1} d x\)

Give an example of:A rational function whose antiderivative is not a rational function.
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