Problem 35 - Calculus Single Variable

Question

Evaluate the given definite integral. \(\int_{1}^{2} \frac{2 x^{2}-x+2}{x\left(x^{2}+1\right)} d x\)

Step-by-Step Solution

Answer

\( \int_{1}^{2} \frac{2 x^{2}-x+2}{xig(x^{2}+1\big)} \, dx = \frac{1}{2}(5\ln(2) - \ln(5)) \).

Step 1: Simplify the Integral Expression

We need to evaluate the integral \(\int_{1}^{2} \frac{2x^{2} - x + 2}{x(x^{2} + 1)} \, dx\). First, simplify the integrand by rewriting it as a sum of simpler fractions. Decompose using partial fraction decomposition:\[ \frac{2x^{2} - x + 2}{x(x^{2} + 1)} = \frac{A}{x} + \frac{Bx + C}{x^{2} + 1} \]Multiply through by the denominator \(x(x^{2} + 1)\) to get:\[2x^2 - x + 2 = A(x^2 + 1) + (Bx + C)x\].

Step 2: Solve for Coefficients A, B, and C

Expand the equation:\[2x^2 - x + 2 = Ax^2 + A + Bx^2 + Cx\]Combine like terms:\[ (A + B)x^2 + Cx + A = 2x^2 - x + 2 \]Equate the coefficients:- \( A + B = 2 \)- \(C = -1\)- \(A = 2\)Solving these, we find \( A = 2, B = 0, C = -1 \). So:\[\frac{2x^2 - x + 2}{x(x^2 + 1)} = \frac{2}{x} + \frac{-x}{x^2 + 1}\].

Step 3: Split the Integral

Use the simplified integrand from Step 2 to split the integral:\[ \int_{1}^{2} \left( \frac{2}{x} + \frac{-x}{x^2 + 1} \right) \, dx = \int_{1}^{2} \frac{2}{x} \, dx + \int_{1}^{2} \frac{-x}{x^2 + 1} \, dx \].

Step 4: Evaluate the First Integral

Evaluate \( \int_{1}^{2} \frac{2}{x} \, dx \):\[ \int_{1}^{2} \frac{2}{x} \, dx = 2 \ln|x| \bigg|_{1}^{2} = 2\ln(2) - 2\ln(1) = 2\ln(2) \].

Step 5: Evaluate the Second Integral

For \( \int_{1}^{2} \frac{-x}{x^2 + 1} \, dx \), use substitution. Let \( u = x^2 + 1 \), then \( du = 2x \, dx \) or \( \frac{1}{2}du = x \, dx \):\[ \int \frac{-x}{x^2 + 1} \, dx = \int \frac{-1}{2} \frac{1}{u} \, du \]This evaluates to:\[ -\frac{1}{2} \ln|u| + C \].Thus:\[ \int_{1}^{2} \frac{-x}{x^2 + 1} \, dx = -\frac{1}{2} \left( \ln(x^2 + 1) \right) \bigg|_{1}^{2} \]\[= -\frac{1}{2}(\ln(5) - \ln(2)) = -\frac{1}{2}\ln(5/2) \].

Step 6: Combine the Results

Combine the results from Steps 4 and 5 to complete the evaluation:\[ 2\ln(2) - \frac{1}{2}\ln\left(\frac{5}{2}\right) \].Simplify further using logarithm properties:\[ 2\ln(2) - \frac{1}{2}\ln\left(\frac{5}{2}\right) = \frac{4}{2}\ln(2) - \frac{1}{2}\ln(5) + \frac{1}{2}\ln(2) \]\[= \frac{5}{2}\ln(2) - \frac{1}{2}\ln(5) = \frac{1}{2}(5\ln(2) - \ln(5)) \].

Key Concepts

Partial Fractions

The concept of partial fractions is crucial when you want to break down a complex fraction into simpler, more manageable parts. This technique is particularly useful when working with integrals, as it allows you to express a rational function as a sum of simpler fractions, often leading to easier integration.
To apply partial fraction decomposition, you typically start with a rational expression where the degree of the numerator is less than the degree of the denominator. In the given problem, we have:

• Numerator: \(2x^2 - x + 2\)
• Denominator: \(x(x^2 + 1)\)

The goal is to express this fraction as a sum of fractions where each component has a simpler structure.
In this specific instance, we decompose the expression into:

• \(\frac{A}{x} + \frac{Bx + C}{x^2 + 1}\)

This requires determining the coefficients \(A\), \(B\), and \(C\) by equating coefficients after expanding the terms, giving us a set of equations. Solving these equations provides the values for these coefficients, which allows us to rewrite the original expression in a much simpler form for integration.

Logarithmic Integration

Logarithmic integration refers to the integration processes where the natural logarithm (ln) function naturally arises. A common scenario is when you encounter the integral \(\int \frac{1}{x} \, dx\), which evaluates to \(\ln |x| + C\).
In the problem at hand, after partial fraction decomposition and splitting the integral, one part becomes \(\int_{1}^{2} \frac{2}{x} \, dx\). The solution to this integral directly involves logarithms:

• \(2 \ln |x|\) evaluated from 1 to 2 yields \(2 \ln 2\) after applying the limits.

Logarithmic functions are particularly helpful due to their properties, which often simplify complex expressions.
For instance, the rules of logarithms help in the final simplification of results by allowing the combination, expansion, or simplification of logarithmic expressions.

U-Substitution

U-substitution is a vital technique in integral calculus that helps simplify the integration process by performing a change of variable. It is especially useful for integrals that resemble a chain rule differentiation.
In the exercise, we encounter this when integrating \(\int_{1}^{2} \frac{-x}{x^2 + 1} \, dx\). To tackle this, we set a substitution:

• Let \(u = x^2 + 1\), thus \(du = 2x \, dx\), or equivalently \(\frac{1}{2} du = x \, dx\).

This substitution transforms the original integral into one involving \(u\), specifically:

• \(-\frac{1}{2} \int \frac{1}{u} \, du\), leading to \(-\frac{1}{2} \ln|u| + C\).

Applying the limits from the original variable to the new one, we evaluate it as:

• \(-\frac{1}{2}(\ln(5) - \ln(2)) = -\frac{1}{2}\ln(\frac{5}{2})\)

U-substitution thus facilitates the transformation of complex integrals into simpler forms by effectively creating an easier path to find the antiderivative.