What is \(\rm \int \frac{dx}{x(x^2 + 1)}\) equal to?
Evaluating Integrals using Partial Fractions
We are asked to evaluate the definite integral given by: \(\rm \int \frac{dx}{x(x^2 + 1)}\).
This is an integral of a rational function. A common technique to solve such integrals is by using partial fraction decomposition. We need to decompose the integrand \(\frac{1}{x(x^2 + 1)}\) into simpler fractions.
Partial Fraction Decomposition Process
We assume the decomposition takes the form:
\[ \frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1} \]
To find the constants A, B, and C, we combine the terms on the right side:
\[ \frac{1}{x(x^2 + 1)} = \frac{A(x^2 + 1) + (Bx + C)x}{x(x^2 + 1)} \]
Equating the numerators, we get:
\[ 1 = A(x^2 + 1) + (Bx + C)x \]
\[ 1 = Ax^2 + A + Bx^2 + Cx \]
Rearranging the terms by powers of \(x\):
\[ 1 = (A + B)x^2 + Cx + A \]
Now, we compare the coefficients of the powers of \(x\) on both sides of the equation. The left side has \(0x^2 + 0x + 1\).
- Coefficient of \(x^2\): \(A + B = 0\)
- Coefficient of \(x\): \(C = 0\)
- Constant term: \(A = 1\)
From \(A = 1\) and \(A + B = 0\), we get \(1 + B = 0\), which means \(B = -1\).
So, the partial fraction decomposition is:
\[ \frac{1}{x(x^2 + 1)} = \frac{1}{x} + \frac{-x + 0}{x^2 + 1} = \frac{1}{x} - \frac{x}{x^2 + 1} \]
Integration of the Decomposed Function
Now we can integrate the decomposed form:
\[ \int \frac{dx}{x(x^2 + 1)} = \int \left( \frac{1}{x} - \frac{x}{x^2 + 1} \right) dx \]
We can integrate each term separately:
\[ \int \frac{1}{x} dx = \ln|x| + C_1 \]
For the second term, \(\int \frac{x}{x^2 + 1} dx\), we can use a substitution. Let \(u = x^2 + 1\). Then the differential \(du\) is \(du = \frac{d}{dx}(x^2 + 1) dx = 2x dx\). This means \(x dx = \frac{1}{2} du\).
Substituting this into the integral:
\[ \int \frac{x}{x^2 + 1} dx = \int \frac{1}{u} \left(\frac{1}{2} du\right) = \frac{1}{2} \int \frac{1}{u} du \]
\[ = \frac{1}{2} \ln|u| + C_2 \]
Substitute back \(u = x^2 + 1\):
\[ = \frac{1}{2} \ln|x^2 + 1| + C_2 \]
Since \(x^2 + 1\) is always positive for real \(x\), \(|x^2 + 1| = x^2 + 1\).
\[ = \frac{1}{2} \ln(x^2 + 1) + C_2 \]
Combining the results for both integrals:
\[ \int \left( \frac{1}{x} - \frac{x}{x^2 + 1} \right) dx = \ln|x| - \frac{1}{2} \ln(x^2 + 1) + C \]
where \(C = C_1 + C_2\) is the constant of integration.
Simplifying the Result using Logarithm Properties
We can rewrite the expression using logarithm properties, specifically \(m \ln a = \ln (a^m)\) and \(\ln a - \ln b = \ln \left(\frac{a}{b}\right)\).
First, rewrite \(\ln|x|\) as \(\frac{1}{2} \cdot 2 \ln|x|\). Since \(2 \ln|x| = \ln(|x|^2) = \ln(x^2)\), we have \(\ln|x| = \frac{1}{2} \ln(x^2)\).
So the expression becomes:
\[ \frac{1}{2} \ln(x^2) - \frac{1}{2} \ln(x^2 + 1) + C \]
Factor out \(\frac{1}{2}\):
\[ \frac{1}{2} \left( \ln(x^2) - \ln(x^2 + 1) \right) + C \]
Using the subtraction property of logarithms:
\[ \frac{1}{2} \ln \left( \frac{x^2}{x^2 + 1} \right) + C \]
This matches one of the given options.
Conclusion
The evaluation of the integral \(\rm \int \frac{dx}{x(x^2 + 1)}\) results in \(\rm \frac{1}{2}ln\left(\frac{x^2}{x^2 + 1}\right) + C\).
Comparing this result with the provided options, we find a match.
Revision Table: Key Concepts for Integration
| Concept | Description | Application in this problem |
|---|---|---|
| Partial Fraction Decomposition | A technique to rewrite rational functions \(\frac{P(x)}{Q(x)}\) as a sum of simpler fractions, where the denominator \(Q(x)\) is factored. | Used to break down \(\frac{1}{x(x^2+1)}\) into \(\frac{A}{x} + \frac{Bx+C}{x^2+1}\). |
| Basic Integration Formulas | Standard rules for integrating common functions, such as \(\int \frac{1}{x} dx = \ln|x|\) and \(\int \frac{1}{u} du = \ln|u|\). | Applied to integrate the decomposed terms \(\frac{1}{x}\) and after substitution. |
| Substitution Method (u-substitution) | A technique to simplify integrals by replacing the variable of integration with a new variable. | Used for the term \(\int \frac{x}{x^2+1} dx\) by setting \(u = x^2+1\). |
| Logarithm Properties | Rules governing logarithmic operations, such as \(\ln a - \ln b = \ln(a/b)\) and \(m \ln a = \ln(a^m)\). | Used to simplify the final logarithmic expression into the form presented in the options. |
| Constant of Integration (C) | An arbitrary constant added to the result of an indefinite integral. | Included in the final answer as required for indefinite integrals. |
Additional Information: Rational Function Integration
Integrals of rational functions, i.e., functions of the form \(\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials, are often solved using the method of partial fraction decomposition. The process involves several steps:
- If the degree of \(P(x)\) is greater than or equal to the degree of \(Q(x)\), perform polynomial long division first. This results in a polynomial plus a proper rational function (where the numerator degree is less than the denominator degree).
- Factor the denominator \(Q(x)\) into linear factors \((ax+b)\) and irreducible quadratic factors \((cx^2+dx+e)\).
- Write the proper rational function as a sum of partial fractions based on the factors of the denominator:
- For each linear factor \((ax+b)\) with multiplicity \(n\), include terms \(\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}\).
- For each irreducible quadratic factor \((cx^2+dx+e)\) with multiplicity \(m\), include terms \(\frac{B_1x+C_1}{cx^2+dx+e} + \frac{B_2x+C_2}{(cx^2+dx+e)^2} + \dots + \frac{B_mx+C_m}{(cx^2+dx+e)^m}\).
- Solve for the unknown constants (A's, B's, C's) by equating the numerators of the original function and the partial fraction decomposition and comparing coefficients or by substituting convenient values for \(x\).
- Integrate each of the resulting partial fractions. These integrals are typically standard forms (like \(\int \frac{1}{u} du\), \(\int \frac{1}{u^2} du\), \(\int \frac{1}{u^2+a^2} du\), \(\int \frac{u}{u^2+a^2} du\)).
In this problem, the denominator was already factored into a linear term \(x\) and an irreducible quadratic term \(x^2+1\), and the numerator degree (0) was less than the denominator degree (3), so no initial division was needed.
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