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: Split the numerator
Write $x=2(x-1)-(x-2)$. Then $\dfrac{x}{(x-1)(x-2)}=\dfrac{2}{x-2}-\dfrac{1}{x-1}$.

Step 2: Integrate each term
$\int\dfrac{2}{x-2}dx=2\log|x-2|$ and $\int\dfrac{1}{x-1}dx=\log|x-1|$.

Step 3: Compare the options
The result $2\log|x-2|-\log|x-1|=\log\dfrac{(x-2)^{2}}{x-1}$ appears only in option D.

Final Answer:
Splitting and integrating gives log((x - 2)^2 / (x - 1)) + c. \[ \boxed{\text{(D) }\log\left(\dfrac{(x-2)^{2}}{x-1}\right)+c} \]