Correct Answer - Option 2 :
\(\rm log \left|{(x-2)\over (x - 1)}\right|+c\)
Concept:
Using partial fraction method
\(\rm 1\over (x-a).(x -b) \) = \(A\over(x-a)\) + \(B\over(x - b)\)
\(\rm \int \frac{{dx}}{x} = \log \left| x \right|\; + c\)
Calculation:
I = \(\int \rm \frac{{dx}}{{(x-2)\;\left( {x - 1} \right)}}\)
⇒ \(\rm 1\over (x-2).(x -1) \) = \(A\over(x-2)\) + \(B\over(x - 1)\)
⇒ 1 = A(x - 1) + B(x - 2)
Compair cofficient both sides
Cofficient of x is A + B = 0
Coffiecient of constant 1 = -A - 2B
Solving the equation we get
A = 1, B = -1
⇒ \(\int \rm \frac{{dx}}{{(x-2)\;\left( {x - 1} \right)}}\) = \(\int {dx\over (x-2)} \) - \(\int {dx\over (x-1)} \)
⇒ log|x - 2| - log|x - 1| + c
= \(\rm log \left|{(x-2)\over (x - 1)}\right|+c\) [∵ log m - log n = log(\(\rm \frac mn\))]
