Integral Calculus Question 21
Question: $ \int_{{}}^{{}}{\frac{x-1}{(x-3)(x-2)}dx=} $
[Roorkee 1978]
Options:
A) $ \log |(x-3)|-\log |(x-2)|+c $
B) $ \log |{{(x-3)}^{2}}|-\log |(x-2)|+c $
C) $ \log |(x-3)|+\log |(x-2)|+c $
D) $ \log |{{(x-3)}^{2}}|+\log |(x-2)|+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{x-1}{(x-3)(x-2)},dx} $
$ =\int_{{}}^{{}}{\frac{x-3}{(x-3)(x-2)},dx+\int_{{}}^{{}}{\frac{2}{(x-3)(x-2)}}},dx $
$ =\log [ \frac{(x-2){{(x-3)}^{2}}}{{{(x-2)}^{2}}} ]+c=\log [ \frac{{{(x-3)}^{2}}}{(x-2)} ]+c. $
Trick : By inspection, $ \frac{d}{dx}{ \log (x-3)-\log (x-2) } $
$ =\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{(x-3)(x-2)} $
$ \Rightarrow \frac{d}{dx}{ 2\log (x-3)-\log (x-2) } $
$ =\frac{2}{x-3}-\frac{1}{x-2}=\frac{x-1}{(x-3)(x-2)} $ .
BETA
