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KCET · Maths · Indefinite Integration

\( \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=A \log |x-1|+B \log |x+2|+C \log |x-3|+K \), then \( A, B \),
C are respectively

  1. A \( \frac{1}{6}, \frac{1}{3}, \frac{1}{5} \)
  2. B \( \frac{-1}{6}, \frac{-1}{3}, \frac{1}{2} \)
  3. C \( \frac{-1}{6}, \frac{1}{3}, \frac{-1}{2} \)
  4. D \( \frac{1}{6}, \frac{-1}{3}, \frac{1}{3} \)
Verified Solution

Answer & Solution

Correct Answer

(B) \( \frac{-1}{6}, \frac{-1}{3}, \frac{1}{2} \)

Step-by-step Solution

Detailed explanation

(B)
\( \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=A \log |x-1|+B \log |x+2|+C \log |x-3|+K \)
Diff w.r.t ' \( x \) '
\( \frac{2 x-1}{(x-1)(x+2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x-3)} \)
\( 2 x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2) \)
Comparing constants \( -1=-6 A+3 B-2 C \)
\( A=\frac{-1}{6}, B=\frac{-1}{3}, C=\frac{1}{2} \)