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Chapter 13: Problem 2

Determine \(\int\left(6 . e^{3 x-5}+\frac{4}{3 x-2}-5^{2 x+1}\right) \mathrm{d} x\).

Short Answer

Expert verified
The short answer is: \(\int\left(6 . e^{3 x-5}+\frac{4}{3 x-2}-5^{2 x+1}\right) \mathrm{d} x = 2e^{3x-5} + \frac{4}{3} \ln |3x-2| - \frac{1}{2\ln 5} 5^{2x+1} + C\)

Step by step solution

01

Identify the types of functions

First, let's identify the types of functions we have in the given expression: 1. Exponential function: \(6e^{3x-5}\) 2. Rational function: \(\frac{4}{3x-2}\) 3. Power function: \(5^{2x+1}\) We will integrate each of these functions separately and then combine the results.
02

Integrate the exponential function

To integrate the exponential function \(6e^{3x-5}\), we will use the following direct integration formula: \(\int e^{ax} dx = \frac{1}{a} e^{ax} + C\) In this case, \(a = 3\), so following the formula: \(\int 6e^{3x-5} dx = \frac{6}{3} e^{3x-5} + C_1 = 2e^{3x-5} + C_1\)
03

Integrate the rational function

To integrate the rational function \(\frac{4}{3x-2}\), we will first notice that the derivative of the denominator is a constant, \(3\). In this case, we will use a direct integration formula: \(\int \frac{1}{ax + b} dx = \frac{1}{a} \ln |ax + b| + C\) In this case, \(a = 3\) and \(b = -2\). Therefore: \(\int \frac{4}{3x-2} dx = \frac{4}{3} \ln |3x-2| + C_2\)
04

Integration of the power function

To integrate the power function \(5^{2x+1}\), we will use the following substitution method: let \(u = 2x+1, du = 2 dx\). So, the power function becomes: \(\int 5^u \frac{1}{2}du\) Now, we will use a direct integration formula for the power function: \(\int a^u du = \frac{1}{\ln a} a^u + C\) In this case, \(a = 5\), and following the formula, we have: \(\int 5^u \frac{1}{2}du = \frac{1}{2\ln 5} 5^u + C_3 = \frac{1}{2\ln 5} 5^{2x+1} + C_3\)
05

Combine the results

Now, we can combine the results of the three integrations: \(\int\left(6 . e^{3 x-5}+\frac{4}{3 x-2}-5^{2 x+1}\right) \mathrm{d} x = 2e^{3x-5} + \frac{4}{3} \ln |3x-2| - \frac{1}{2\ln 5} 5^{2x+1} + C\) So, the final answer is: \(\int\left(6 . e^{3 x-5}+\frac{4}{3 x-2}-5^{2 x+1}\right) \mathrm{d} x = 2e^{3x-5} + \frac{4}{3} \ln |3x-2| - \frac{1}{2\ln 5} 5^{2x+1} + C\)

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Key Concepts

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

Exponential Functions
An exponential function typically looks like this:
  • Function: \(e^{ax}\)
  • This function rapidly increases or decreases, depending on the sign of \(a\).
  • The base, \(e\), is known as Euler's number, approximately equal to 2.718.
In integration, we handle exponential functions by recognizing that they have a straightforward integration formula.
For exponential functions of the form \(e^{ax}\), the integration rule is simple:\[\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\]Here, \(C\) is the constant of integration. This formula works perfectly since the derivative of \(e^{ax}\) is simply \(a \, e^{ax}\), reflecting back to its original form with a constant multiplier.
This method ensures smooth integration without complex transformations or substitutions, making it a favorite for its simplicity when working with such functions.
Rational Functions
Rational functions involve fractions where the numerator and the denominator are polynomials. An example is \(\frac{1}{ax + b}\), where such functions often require a specific approach to integrate:
  • Function form: \(\frac{c}{ax+b}\)
  • Calculate the derivative of the denominator to assist in integration.
  • Use logarithmic integration for such cases.
The integral of a rational function of the form \(\frac{1}{ax+b}\) can be directly solved with:\[\int \frac{1}{ax+b}\,dx = \frac{1}{a} \ln|ax+b| + C\]In rational functions, integration leverages the natural log function (\(\ln\)), especially useful if you notice that the denominator's derivative is a constant or relevant for integration.
Recognizing such patterns not only makes integration simpler but also more logical in its process and execution.
Integration by Substitution
Integration by substitution is a common technique used to simplify complex integrals. It works by finding a substitute for a part of the integrand:
  • It simplifies the original function into a basic integrable format.
  • Typically involves letting \(u\) be a function of \(x\).
  • Don't forget to adjust \(du\) for the entire integrand.
The rule revolves around converting expressions within the integral, making them easier to manage:\[\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du\]This method brings integration into a form which is straightforward to integrate, followed by substituting back to the original variable.
For instance, if faced with integrals like \(5^{2x+1}\):
  • Let \(u = 2x+1\), implying \(du = 2 \, dx\).
  • Integral becomes \(\int 5^u \cdot \frac{1}{2} \, du\).
This illustrates how substitution helps reduce complexity by focusing on building blocks of the integrand, making integration manageable and efficient.

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Most popular questions from this chapter

Determine the area bounded by the curve \(y=f(x)\), the \(x\)-axis and the stated ordinates in the following cases: (a) \(y=x^{2}-3 x+4, \quad x=0\) and \(x=4\) (b) \(y=3 x^{2}+5, \quad x=-2\) and \(x=3\) (c) \(y=5+6 x-3 x^{2}, x=0\) and \(x=2\) (d) \(y=-3 x^{2}+12 x+10, \quad x=1\) and \(x=4\) (e) \(y=x^{3}+10, \quad x=-1\) and \(x=2\)

(a) Determine \(\int\left(8 x^{3}+6 x^{2}-5 x+4\right) \mathrm{d} x\). (b) If \(I=\int\left(4 x^{3}-3 x^{2}+6 x-2\right) \mathrm{d} x\), determine the value of \(I\) when \(x=4\), given that when \(x=2, I=20\).

(a) Evaluate the area under the curve \(y=5 . e^{x}\) between \(x=0\) and \(x=3\). (b) Show that the straight line \(y=-2 x+28\) crosses the curve \(y=36-x^{2}\) at \(x=-2\) and \(x=4\). Hence, determine the area enclosed by the curve \(y=36-x^{2}\) and the line \(y=-2 x+28\)

Determine: (a) \(\int 2 \sin (3 x+1) \mathrm{d} x\) (e) \(\int 4^{2 x-3} \mathrm{~d} x\) (b) \(\int \sqrt{5-2 x} \mathrm{~d} x\) (f) \(\int 6 \cos (1-2 x) \mathrm{d} x\) (c) \(\int 6 e^{1-3 x} \mathrm{~d} x\) (g) \(\int \frac{5}{3 x-2} d x\) (d) \(\int(4 x+1)^{3} \mathrm{~d} x\) (h) \(\int 3 \sec ^{2}(1+4 x) \mathrm{d} x\)

Express \(\frac{2 x^{2}+14 x+10}{2 x^{2}+9 x+4}\) in partial fractions and hence determine $$ \int \frac{2 x^{2}+14 x+10}{2 x^{2}+9 x+4} \mathrm{~d} x $$
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