Question Statement

Evaluate the definite integral:

${\int }_2^5\frac{1}{x(x+1)}dx$

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Background and Explanation

This problem requires partial fraction decomposition to rewrite the rational function as a difference of simpler fractions, which can then be integrated using the natural logarithm rule. The key is recognizing that $\frac{1}{x(x+1)}$ can be split into terms with linear denominators.

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Solution

We begin with the partial fraction decomposition already applied to the integrand:

$\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$

Now we evaluate the definite integral by splitting it into two separate integrals:

$\begin{array}{rl}{\int }_2^5\frac{1}{x(x+1)}dx& ={\int }_2^5\frac{1}{x}dx-{\int }_2^5\frac{1}{x+1}dx\\ & ={\left[\ln x\right]}_2^5-{\left[\ln (x+1)\right]}_2^5\\ & =(\ln 5-\ln 2)-(\ln (5+1)-\ln (2+1))\\ & =(\ln 5-\ln 2)-(\ln 6-\ln 3)\\ & =\ln 5-\ln 2-\ln 6+\ln 3\\ & =\ln \left(\frac{5\times 3}{2\times 6}\right)\\ & =\ln \left(\frac{15}{12}\right)\\ & =\ln \left(\frac{5}{4}\right)\end{array}$

Reasoning:

• First, we apply the partial fraction decomposition to break the complex fraction into two simpler terms that we can integrate separately.
• We integrate $\frac{1}{x}$ to get $\ln x$ and $\frac{1}{x+1}$ to get $\ln (x+1)$ (the derivative of the denominator matches the numerator structure).
• The notation ${\left[\ln x\right]}_2^5$ indicates we evaluate $\ln x$ at $x=5$ and subtract its value at $x=2$.
• After substituting the bounds, we apply logarithm properties: $\ln a-\ln b=\ln \frac{a}{b}$ to combine terms, and $\ln a+\ln b=\ln (ab)$ to group the positive and negative terms separately.
• Finally, we simplify the fraction $\frac{15}{12}$ to $\frac{5}{4}$.

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Key Formulas or Methods Used

• Partial Fraction Decomposition: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$
• Logarithmic Integration: $\int \frac{1}{x}dx=\ln |x|+C$ and $\int \frac{1}{x+a}dx=\ln |x+a|+C$
• Fundamental Theorem of Calculus: ${\int }_a^{b}f(x)dx=F(b)-F(a)$ where $F$ is the antiderivative of $f$
• Logarithm Properties:
  • $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$
  • $\ln a+\ln b=\ln (ab)$

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Summary of Steps

1. Decompose the integrand using partial fractions: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$
2. Split the integral into two simpler integrals: $\int \frac{1}{x}dx-\int \frac{1}{x+1}dx$
3. Integrate both terms to obtain natural logarithms: ${\left[\ln x\right]}_2^5-{\left[\ln (x+1)\right]}_2^5$
4. Evaluate at the upper bound (5) and lower bound (2) for both terms
5. Simplify the resulting expression using logarithm rules to combine terms into a single logarithm
6. Reduce the final fraction to obtain $\ln \left(\frac{5}{4}\right)$