Question

\[ \int \frac{x \mathrm{~d} x}{(x-1)(x-2)}= \]

• A. \(\mathrm{log}(\frac{x-1}{x-2})+\mathrm{c}\), where c is the constant of integration
• B. \(\mathrm{log}(\frac{x-2}{(x-1)^{2}})+\mathrm{c}\), where c is the constant of integration
• C. \(\mathrm{log}(\frac{x-2}{x-1})+\mathrm{c}\), where c is the constant of integration
• D. \(\mathrm{log}(\frac{(x-2)^{2}}{x-1})+\mathrm{c}\), where c is the constant of integration

Hint:

Resolve x/((x-1)(x-2)) into partial fractions and integrate each term.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The integrand is a proper rational function with two distinct linear factors, so use partial fractions.

Step 2: Partial fractions
Let \(\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}\). Then \(x=A(x-2)+B(x-1)\).
Put \(x=1\): \(A=-1\). Put \(x=2\): \(B=2\).

Step 3: Integrate
\(\int\left(\dfrac{-1}{x-1}+\dfrac{2}{x-2}\right)dx=-\log|x-1|+2\log|x-2|+c\).

Step 4: Combine the logarithms
\(=\log\dfrac{(x-2)^{2}}{x-1}+c\).

Final Answer:
The integral is log((x - 2)^2 / (x - 1)) + c. \[ \boxed{\text{(D) }\log\left(\dfrac{(x-2)^{2}}{x-1}\right)+c} \]