Chapter 4: Problem 50
Evaluate. $$\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x$$
Short Answer
Expert verified
The integral evaluates to \(\ln(4.5).\)
Step by step solution
01
Simplify the integrand
First, simplify the function inside the integral. Observe that \(\frac{2x + 3}{x^2 + 3x}\) can be simplified by partial fraction decomposition. The expression \(x^2 + 3x\) can be factored as \(x(x + 3).\)
02
Set up partial fractions
Set up the partial fraction decomposition: \(\frac{2x + 3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}.\) Solve for A and B by multiplying through by the common denominator \(x(x + 3).\)
03
Solve for constants A and B
Solve the equation \(2x + 3 = A(x + 3) + Bx.\) Matching coefficients, get the system: \(A + B = 2\) and \(3A = 3.\) Solving these, A = 1 and B = 1.
04
Integrate each term separately
Now, rewrite the integral as \(\int_{1}^{3} \frac{1}{x} dx + \int_{1}^{3} \frac{1}{x + 3} dx.\) Integrate each term separately: \(\int_{1}^{3} \frac{1}{x} dx = \ln|x|\bigg|_{1}^{3},\) and \(\int_{1}^{3} \frac{1}{x + 3} dx = \ln|x + 3|\bigg|_{1}^{3}.\)
05
Evaluate the definite integrals
Evaluate the definite integrals: \(\ln|x|\bigg|_{1}^{3} = \ln(3) - \ln(1) = \ln(3),\) and \(\ln|x + 3|\bigg|_{1}^{3} = \ln(6) - \ln(4).\)
06
Combine the results
Combine the results: \(\ln(3) + (\ln(6) - \ln(4)) = \ln(3) + \ln(\frac{6}{4}) = \ln(3) + \ln(1.5).\)
07
Simplify the final expression
Simplify the final expression: \(\ln(3) + \ln(1.5) = \ln(3 \times 1.5) = \ln(4.5).\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
In integral calculus, a definite integral is an integral with upper and lower limits. It computes the net area under a curve within a specified range on the x-axis, often between two points.
The integral \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\) guides us through such a calculation, bounded between 1 and 3.
Unlike indefinite integrals which result in a function plus a constant, definite integrals yield a specific value. To evaluate, apply the Fundamental Theorem of Calculus: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] where F(x) is the antiderivative of f(x). This theorem makes definite integrals instrumental in practical applications across physics, engineering, and beyond.
The integral \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\) guides us through such a calculation, bounded between 1 and 3.
Unlike indefinite integrals which result in a function plus a constant, definite integrals yield a specific value. To evaluate, apply the Fundamental Theorem of Calculus: \[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \] where F(x) is the antiderivative of f(x). This theorem makes definite integrals instrumental in practical applications across physics, engineering, and beyond.
Partial Fraction Decomposition
Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions.
It's particularly useful when integrating fractions.
Consider the integral \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\). Here, \(\frac{2x + 3}{x^2 + 3x}\) is decomposed into partial fractions:
\[ \frac{2x + 3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} \]
By solving for constants A and B through the method of equating coefficients, we can simplify the integral further.
In our example, setting up and solving \((2x + 3 = A(x + 3) + Bx)\) leads to \(A = 1\) and \(B = 1\).These simpler fractions \(\frac{1}{x}\) and \(\frac{1}{x + 3}\) are then much easier to integrate.
It's particularly useful when integrating fractions.
Consider the integral \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\). Here, \(\frac{2x + 3}{x^2 + 3x}\) is decomposed into partial fractions:
\[ \frac{2x + 3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} \]
By solving for constants A and B through the method of equating coefficients, we can simplify the integral further.
In our example, setting up and solving \((2x + 3 = A(x + 3) + Bx)\) leads to \(A = 1\) and \(B = 1\).These simpler fractions \(\frac{1}{x}\) and \(\frac{1}{x + 3}\) are then much easier to integrate.
Logarithmic Integration
Logarithmic integration, involving the natural logarithm function, often appears when integrating functions of the form \(\int \frac{1}{x} dx\).
For example, after decomposing \(\frac{2x + 3}{x^2 + 3x}\) via partial fractions, we separate it into integrals like \(\int_{1}^{3} \frac{1}{x} dx\).
Integrating \(\frac{1}{x}\) and \(\frac{1}{x+3}\) results in logarithmic functions:
\[ \int \frac{1}{x} \, dx = \ln|x| + C \] \[ \int \frac{1}{x + 3} \, dx = \ln|x + 3| + C \]
Using the Fundamental Theorem of Calculus, we evaluate these integrals from 1 to 3.
This process results in expressions involving \( \ln(3) \) and \( \ln(6) - \ln(4)\), combining to give the final result.
For example, after decomposing \(\frac{2x + 3}{x^2 + 3x}\) via partial fractions, we separate it into integrals like \(\int_{1}^{3} \frac{1}{x} dx\).
Integrating \(\frac{1}{x}\) and \(\frac{1}{x+3}\) results in logarithmic functions:
\[ \int \frac{1}{x} \, dx = \ln|x| + C \] \[ \int \frac{1}{x + 3} \, dx = \ln|x + 3| + C \]
Using the Fundamental Theorem of Calculus, we evaluate these integrals from 1 to 3.
This process results in expressions involving \( \ln(3) \) and \( \ln(6) - \ln(4)\), combining to give the final result.
Simplification of Integrals
Simplifying integrals often involves rewriting the integral in a more manageable form.
For \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\), factorize and decompose to break down into simpler integrals.
Following partial fraction decomposition, the integral becomes:
\[ \int_{1}^{3} \frac{1}{x} dx + \int_{1}^{3} \frac{1}{x + 3} dx \]
These integrals lead to the natural logarithms: \(\ln|x|\bigg|_{1}^{3}\) and \(\ln|x + 3|\bigg|_{1}^{3}\).
Use properties of logarithms to combine results:
This combination simplifies integration results, presenting the solution in the simplest form. Always aim to reduce to the most compact and understandable solution.
For \(\int_{1}^{3} \frac{2 x+3}{x^{2}+3 x} d x\), factorize and decompose to break down into simpler integrals.
Following partial fraction decomposition, the integral becomes:
\[ \int_{1}^{3} \frac{1}{x} dx + \int_{1}^{3} \frac{1}{x + 3} dx \]
These integrals lead to the natural logarithms: \(\ln|x|\bigg|_{1}^{3}\) and \(\ln|x + 3|\bigg|_{1}^{3}\).
Use properties of logarithms to combine results:
- \[ \ln(3) \- \ln(1) = \ln(3) \] and \[ \ln(6) \- \ln(4)\]
- \[ \ln(3 \times \frac{6}{4}) = \ln(4.5)\]
This combination simplifies integration results, presenting the solution in the simplest form. Always aim to reduce to the most compact and understandable solution.