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\( \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=A \log |x-1|+B \log |x+2|+C \log |x-3|+K \), then \( A, B \),
C are respectively
Options:
C are respectively
Solution:
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Verified Answer
The correct answer is:
\( \frac{-1}{6}, \frac{-1}{3}, \frac{1}{2} \)
(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 \( -1=-6 A+3 B-2 C \)
\( A=\frac{-1}{6}, B=\frac{-1}{3}, C=\frac{1}{2} \)
\( \int \frac{2 x-1}{(x-1)(x+2)(x-3)} d x=A \log |x-1|+B \log |x+2|+C \log |x-3|+K \)
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 \( -1=-6 A+3 B-2 C \)
\( A=\frac{-1}{6}, B=\frac{-1}{3}, C=\frac{1}{2} \)
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