Question

Evaluate the definite integral: $$\int_{5}^{10} \frac{dx}{(x-1)(x-2)}$$

• A. $\log\left|\frac{27}{32}\right|$
• B. $\log\left|\frac{3}{4}\right|$
• C. $\log\left|\frac{8}{9}\right|$
• D. $\log\left|\frac{32}{27}\right|$

Hint:

For partial fractions where factors differ by exactly 1 unit, the numerator splits instantly as: $\frac{1}{\text{Small} \cdot \text{Large}} = \frac{1}{\text{Small}} - \frac{1}{\text{Large}}$.
Here, $\frac{1}{(x-2)(x-1)} = \frac{1}{x-2} - \frac{1}{x-1}$. Working out the boundaries cleanly inside the consolidated log format ($\frac{x-2}{x-1}$) eliminates sign errors completely.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires calculating the value of a definite integral with a rational integrand over the interval $[5, 10]$.

Step 2: Key Formula or Approach:
We resolve the proper algebraic fraction using the method of partial fractions:
$$\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$$ By inspection or standard clearing of denominators, we observe:
$$\frac{1}{(x-1)(x-2)} = \frac{1}{x-2} - \frac{1}{x-1}$$ Then we apply the fundamental theorem of calculus: $\int \frac{1}{x-a}\ dx = \log|x-a|$.

Step 3: Detailed Explanation:
Let the given definite integral be $I$:
$$I = \int_{5}^{10} \left[ \frac{1}{x-2} - \frac{1}{x-1} \right]\ dx$$ Integrating each term yields:
$$I = \Big[ \log|x-2| - \log|x-1| \Big]_{5}^{10}$$ Using logarithm properties, combine the terms before inserting bounds:
$$I = \left[ \log\left|\frac{x-2}{x-1}\right| \right]_{5}^{10}$$ Substitute the upper limit ($x = 10$):
$$\text{Upper Value} = \log\left|\frac{10-2}{10-1}\right| = \log\left|\frac{8}{9}\right|$$ Substitute the lower limit ($x = 5$):
$$\text{Lower Value} = \log\left|\frac{5-2}{5-1}\right| = \log\left|\frac{3}{4}\right|$$ Subtract the lower evaluation from the upper evaluation:
$$I = \log\left|\frac{8}{9}\right| - \log\left|\frac{3}{4}\right|$$ $$I = \log\left| \frac{8}{9} \times \frac{4}{3} \right| = \log\left|\frac{32}{27}\right|$$

Step 4: Final Answer:
The value of the definite integral is $\log\left|\frac{32}{27}\right|$, which corresponds to option (D).