The value of integral \( \mathrm{I}=\int_{3}^{5}\displaystyle\frac{x^2}{(x-1)(x-2)}\,dx \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2-\log 2+4\log 3 \)
The numerator and denominator have the same degree, so we divide first. Since \( (x-1)(x-2)=x^2-3x+2 \),
\( \displaystyle \frac{x^2}{(x-1)(x-2)}=1+\frac{3x-2}{(x-1)(x-2)} \)
The proper fraction splits into partial fractions:
\( \displaystyle \frac{3x-2}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2} \)
\( \displaystyle x=1:\ A=\frac{3-2}{1-2}=-1 \)
\( \displaystyle x=2:\ B=\frac{6-2}{2-1}=4 \)
(The values of A and B come from covering up the matching factor and putting in its root.)
\( \displaystyle \mathrm{I}=\int_{3}^{5}\left(1-\frac{1}{x-1}+\frac{4}{x-2}\right)dx \)
\( \displaystyle =\Big[x-\log|x-1|+4\log|x-2|\Big]_{3}^{5} \)
\( \displaystyle =(5-\log 4+4\log 3)-(3-\log 2+4\log 1) \)
Using \( \log 4=2\log 2 \) and \( \log 1=0 \):
\( \displaystyle =2-2\log 2+\log 2+4\log 3 \)
\( \displaystyle =2-\log 2+4\log 3 \)
Why the other options are incorrect
Option (2) \( \log 2+4\log 3 \) — Incorrect. The integral of the constant 1 from 3 to 5 gives 2, which is missing here.
Option (3) \( 3-\log 2+\log 3 \) — Incorrect. The constant part should be 5 − 3 = 2, and the coefficient 4 of \( \log 3 \) is lost.
Option (4) \( 2-\log 2+\log 3 \) — Incorrect. The coefficient 4 from \( \frac{4}{x-2} \) has been dropped; \( 4\log|x-2| \) gives \( 4\log 3 \), not \( \log 3 \).