Question

$\int \frac{x dx}{(x-1)(x-2)} =$}

• A. $\log \left( \frac{x-1}{x-2} \right) + c$
• B. $\log \left( \frac{x-2}{(x-1)^2} \right) + c$
• C. $\log \left( \frac{x-2}{x-1} \right) + c$
• D. $\log \left( \frac{(x-2)^2}{x-1} \right) + c$

Hint:

$\log A - \log B = \log(A/B)$ and $n \log A = \log(A^n)$.

Answer 1

The Correct Option is D

Step 1: Concept
Use partial fraction decomposition: $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.

Step 2: Meaning
$x = A(x-2) + B(x-1)$.
If $x=1$, $1 = A(-1) \implies A = -1$.
If $x=2$, $2 = B(1) \implies B = 2$.

Step 3: Analysis
Integral becomes: $\int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx = -\log(x-1) + 2\log(x-2) + c$.

Step 4: Conclusion
Using log properties: $\log(x-2)^2 - \log(x-1) = \log \left( \frac{(x-2)^2}{x-1} \right) + c$. Final Answer: (D)