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: Look at the integral.
We need $\displaystyle \int_{5}^{10} \frac{dx}{(x-1)(x-2)}$. The bottom is a product, so we split it using partial fractions.
Step 2: Break into simple fractions.
Write $\dfrac{1}{(x-1)(x-2)} = \dfrac{A}{x-1} + \dfrac{B}{x-2}$. Solving gives $A = -1$, $B = 1$, so \[ \frac{1}{(x-1)(x-2)} = \frac{1}{x-2} - \frac{1}{x-1} \]
Step 3: Integrate each piece.
Using $\int \frac{1}{x-a}\,dx = \ln|x-a|$: \[ I = \Big[\ln|x-2| - \ln|x-1|\Big]_{5}^{10} = \left[\ln\left|\frac{x-2}{x-1}\right|\right]_{5}^{10} \]
Step 4: Put in the upper limit.
At $x = 10$: $\ln\left|\dfrac{8}{9}\right|$.
Step 5: Put in the lower limit.
At $x = 5$: $\ln\left|\dfrac{3}{4}\right|$.
Step 6: Subtract and combine.
\[ I = \ln\frac{8}{9} - \ln\frac{3}{4} = \ln\left(\frac{8}{9} \times \frac{4}{3}\right) = \ln\frac{32}{27} \] \[ \boxed{\log\left|\dfrac{32}{27}\right|} \]