What is the value of k such that integration of \(\rm \dfrac{3x^2+8-4k}{x}\) with respect to x, may be a rational function ?
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Understanding the Problem: Rational Functions and Integration
The question asks for the value of a constant, \(k\), such that the integral of the given expression, \(\rm \dfrac{3x^2+8-4k}{x}\), results in a rational function. Let's first understand what a rational function is and how integration affects it.
A rational function is a function that can be expressed as the fraction \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \) is not the zero polynomial.
When we integrate a polynomial \( x^n \) where \( n \neq -1 \), the result is \( \frac{x^{n+1}}{n+1} \) (plus a constant of integration). This result is also a rational function (specifically, a polynomial if \(n+1\) is a non-negative integer). However, the integral of \( x^{-1} \) or \( \frac{1}{x} \) is \( \ln|x| \) (plus a constant), which is not a rational function because it involves a logarithm.
Step-by-Step Integration and Analysis
Let's integrate the given expression: \( \rm \int \dfrac{3x^2+8-4k}{x} \, dx \).
First, we can simplify the integrand by dividing each term in the numerator by \( x \):
\( \dfrac{3x^2+8-4k}{x} = \dfrac{3x^2}{x} + \dfrac{8-4k}{x} = 3x + \dfrac{8-4k}{x} \)
Now, we can integrate term by term:
\( \int \left( 3x + \dfrac{8-4k}{x} \right) \, dx = \int 3x \, dx + \int \dfrac{8-4k}{x} \, dx \)
Let's integrate the first term:
\( \int 3x \, dx = 3 \int x^1 \, dx = 3 \cdot \dfrac{x^{1+1}}{1+1} + C_1 = 3 \cdot \dfrac{x^2}{2} + C_1 = \dfrac{3}{2}x^2 + C_1 \)
This term, \( \dfrac{3}{2}x^2 \), is a rational function (it's a polynomial). This part doesn't cause any issues for the integral being a rational function.
Now, let's integrate the second term. Notice that \( (8-4k) \) is a constant. We can pull it out of the integral:
\( \int \dfrac{8-4k}{x} \, dx = (8-4k) \int \dfrac{1}{x} \, dx \)
The integral of \( \frac{1}{x} \) is \( \ln|x| \). So, the second term integrates to:
\( (8-4k) \int \dfrac{1}{x} \, dx = (8-4k) \ln|x| + C_2 \)
Combining the results from both terms, the complete integral is:
\( \int \dfrac{3x^2+8-4k}{x} \, dx = \dfrac{3}{2}x^2 + (8-4k) \ln|x| + C \)
where \( C = C_1 + C_2 \) is the constant of integration.
Ensuring the Integral is a Rational Function
For the entire resulting function \( \dfrac{3}{2}x^2 + (8-4k) \ln|x| + C \) to be a rational function, the term involving \( \ln|x| \) must disappear. The only way for this term to disappear is if its coefficient, \( (8-4k) \), is equal to zero.
So, we set the coefficient to zero and solve for \( k \):
\( 8 - 4k = 0 \)
\( 8 = 4k \)
Divide both sides by 4:
\( k = \dfrac{8}{4} \)
\( k = 2 \)
Thus, when \( k=2 \), the integral becomes:
\( \int \dfrac{3x^2+8-4(2)}{x} \, dx = \int \dfrac{3x^2+8-8}{x} \, dx = \int \dfrac{3x^2}{x} \, dx = \int 3x \, dx \)
And the integral of \( 3x \) is \( \dfrac{3}{2}x^2 + C \), which is indeed a rational function.
Conclusion on the Value of k
To ensure the integration of \( \rm \dfrac{3x^2+8-4k}{x} \) results in a rational function, the coefficient of the \( \frac{1}{x} \) term in the expanded integrand must be zero. This leads to the equation \( 8 - 4k = 0 \), which yields \( k=2 \).
| Step | Action | Result |
|---|---|---|
| 1 | Simplify the integrand | \( 3x + \dfrac{8-4k}{x} \) |
| 2 | Integrate the expression | \( \int 3x \, dx + \int \dfrac{8-4k}{x} \, dx \) |
| 3 | Evaluate integral of \(3x\) | \( \dfrac{3}{2}x^2 + C_1 \) (Rational) |
| 4 | Evaluate integral of \( \dfrac{8-4k}{x} \) | \( (8-4k) \ln|x| + C_2 \) (Involves non-rational \( \ln|x| \) unless \( 8-4k=0 \)) |
| 5 | Set coefficient of \( \ln|x| \) to zero | \( 8 - 4k = 0 \) |
| 6 | Solve for \(k\) | \( k = 2 \) |
Revision Table: Key Concepts
| Concept | Definition/Description | Relevance to Problem |
|---|---|---|
| Rational Function | A function expressed as a ratio of two polynomials, \(P(x)/Q(x)\). | The desired outcome of the integration. |
| Integration | The process of finding the antiderivative of a function. | The core operation performed on the given expression. |
| Integral of \(x^n\) (\(n \neq -1\)) | \( \dfrac{x^{n+1}}{n+1} + C \). Results in a rational function. | Applies to the \(3x\) term after simplification. |
| Integral of \(1/x\) (\(x^{-1}\)) | \( \ln|x| + C \). Does NOT result in a rational function. | This term must be eliminated from the integral result. |
| Constant \(k\) | A variable in the integrand that affects the coefficient of the \(1/x\) term. | Its value is determined by the condition that the integral is rational. |
Additional Information: Types of Functions
Understanding different types of functions is crucial in calculus. Here are a few types related to this problem:
- Polynomial Functions: Functions of the form \( a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \), where \(a_i\) are constants and \(n\) is a non-negative integer. These are a subset of rational functions.
- Rational Functions: Functions that are ratios of two polynomials, \( P(x)/Q(x) \). Polynomials themselves are rational functions (where \(Q(x)=1\)).
- Transcendental Functions: Functions that are not algebraic (like polynomials or rational functions). Examples include logarithmic functions (like \( \ln x \)), exponential functions (like \( e^x \)), trigonometric functions (like \( \sin x \)), and inverse trigonometric functions. The presence of a transcendental function like \( \ln|x| \) prevents the total function from being rational.
The problem highlights that while integrating polynomials usually results in polynomials (which are rational), integrating terms like \( \frac{1}{x} \) introduces non-rational functions like logarithms. Therefore, to keep the integral rational, any term that integrates to a non-rational function must have a zero coefficient.
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