\(\int\displaystyle\frac{x^2+2}{x^2-3x+2}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x-3\log|x-1|+6\log|x-2|+C\)
Numerator and denominator have the same degree, so the fraction is improper; dividing first:
\( \displaystyle \frac{x^2+2}{x^2-3x+2} = 1+\frac{3x}{(x-1)(x-2)} \)
By partial fractions:
\( 3x = A(x-2)+B(x-1) \)
Putting x = 1 and x = 2:
\( 3 = -A,\quad A = -3 \)
\( 6 = B \)
So
\( \displaystyle \int\frac{x^2+2}{x^2-3x+2}\,dx = \int\left(1-\frac{3}{x-1}+\frac{6}{x-2}\right)dx \)
\( = x-3\log|x-1|+6\log|x-2|+C \)
Why the other options are incorrect
Option (1) \(x+3\log|x-1|-6\log|x-2|+C\) — Incorrect. The signs of A and B are reversed.
Option (2) \(-3\log|x-1|+6\log|x-2|+C\) — Incorrect. It leaves out the term x that comes from the division.
Option (4) \(x-3\log|x-1|+4\log|x-2|+C\) — Incorrect. At x = 2, \(3x = 6\), so B = 6, not 4.