/*! 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 57 Evaluate each integral by first ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 57

Evaluate each integral by first modifying the form of the integrand and then making an appropriate substitution, if needed. $$\int \frac{t+1}{t} d t$$

Short Answer

Expert verified
\( t + \ln |t| + C \)

Step by step solution

01

Simplify the Integrand

Start by splitting the integrand into separate fractions. Rewrite \( \frac{t+1}{t} \) as \( \frac{t}{t} + \frac{1}{t} \). Simplifying further, this becomes \( 1 + \frac{1}{t} \).
02

Integrate Each Term Separately

Next, integrate the simplified expression. The integral becomes \( \int (1 + \frac{1}{t}) \, dt = \int 1 \, dt + \int \frac{1}{t} \, dt \).
03

Evaluate the Integral of 1

The integral of \(1\) with respect to \(t\) is straightforward: \( \int 1 \, dt = t \).
04

Evaluate the Integral of \(\frac{1}{t}\)

The integral of \( \frac{1}{t} \) with respect to \( t \) is the natural logarithm function: \( \int \frac{1}{t} \, dt = \ln |t| \).
05

Combine the Results

Add the results from Steps 3 and 4. The integral \( \int \frac{t+1}{t} \, dt \) simplifies to \( t + \ln |t| + C \), where \( C \) is the constant of integration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Integration Techniques
Integration is a fundamental concept in calculus, used to find areas under curves, among other applications. Various techniques can simplify the process of finding integrals.
One such technique involves simplifying the form of the integrand. Simply put, the integrand is the expression inside the integral sign. By breaking down or altering its form, you can reduce complexity and make the integration process more straightforward.
  • Splitting fractions is a commonly used technique. It involves rewriting a complex fraction into simpler parts. This was done in the example where \( \frac{t+1}{t} \) was split into \( \frac{t}{t} + \frac{1}{t} \).
  • Another method is to factor expressions, which can help when dealing with polynomials or trigonometric functions.
  • Recognizing patterns can also aid in simplifying integrals. If an integrand is similar to a derivative you know, you can use that knowledge to integrate more easily.
By using these techniques, the process of finding integrals becomes more efficient and less prone to errors.
Substitution Method
The substitution method is a powerful tool in integral calculus, often used when the integrand is complex. It involves changing the integral into a simpler form that is easier to evaluate. The process is akin to reversing the chain rule for derivatives.
To perform substitution, you:
  • Select a part of the integrand to substitute, simplifying the integral.
  • Replace this part with a new variable, usually \( u \), and find \( du \), the derivative of \( u \) with respect to \( x \).
  • Rewrite the integral in terms of \( u \) and \( du \).
  • Integrate in the new variable. Often, the integration becomes straightforward.
  • Substitute back the original variable at the end of the process.
Let's consider a simple example: to integrate \( \int (2x \cdot e^{x^2}) dx \), set \( u = x^2 \) making \( du = 2x \, dx \). Then, the integral becomes \( \int e^u \, du \), which is straightforward to integrate.
Natural Logarithm Integration
The natural logarithm, denoted as \( \ln \), plays a crucial role in integral calculus, especially when integrating functions of the form \( \frac{1}{x} \). The integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln |x| + C \), where \( C \) is the constant of integration.
This result stems from the derivative of the natural logarithm function. Recall that \( \frac{d}{dx} \ln|x| = \frac{1}{x} \). Consequently, reversing this operation through integration naturally leads to the logarithm.
Key points to remember include:
  • The natural logarithm integral applies to any term in the integrand in the form of \( \frac{1}{x} \).
  • The result includes the absolute value of \( x \), as the domain of the logarithm excludes negative numbers or zero.
  • Understanding this integration is important because it's a foundational concept that applies directly to many real-world scenarios involving rates of change and growth.
Mastery of natural logarithm integration is essential for tackling a wide range of calculus problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Define \(F(x)\) by \(F(x)=\int_{1}^{x}\left(3 t^{2}-3\right) d t\) (a) Use Part 2 of the Fundamental Theorem of Calculus to find \(F^{\prime}(x)\) (b) Check the result in part (a) by first integrating and then differentiating.

Solve the initial-value problems. $$\frac{d y}{d x}=2+\sin 3 x, y(\pi / 3)=0$$

(a) Show that $$\begin{aligned} \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} &=\frac{n}{n+1} \\ \left[\text { Hint: } \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\right] \end{aligned}$$ (b) Use the result in part (a) to find $$ \lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{1}{k(k+1)} $$

$$\begin{aligned} \sum_{k=1}^{4}\left[(k+1)^{3}-k^{3}\right]=&\left[5^{3}-4^{3}\right]+\left[4^{3}-3^{3}\right] \\\ &+\left[3^{3}-2^{3}\right]+\left[2^{3}-1^{3}\right] \\ =& 5^{3}-1^{3}=124 \end{aligned}$$ For convenience, the terms are listed in reverse order. Note how cancellation allows the entire sum to collapse like a telescope. A sum is said to telescope when part of each term cancels part of an adjacent term, leaving only portions of the first and last terms uncanceled. Evaluate the telescoping sums in these exercises. $$\sum_{k=5}^{17}\left(3^{k}-3^{k-1}\right)$$

(a) The accompanying table shows the fraction of the Moon that is illuminated (as seen from Earth) at midnight (Eastern Standard Time) for the first week of 2005. Find the average fraction of the Moon illuminated during the first week of \(2005 .\) (b) The function \(f(x)=0.5+0.5 \sin (0.213 x+2.481)\) models data for illumination of the Moon for the first 60 days of \(2005 .\) Find the average value of this illumination function over the interval [0,7]. $$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text { DAY } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\\\hline \text { ILLUMINATION } & 0.74 & 0.65 & 0.56 & 0.45 & 0.35 & 0.25 & 0.16 \\\\\hline\end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.