Explanation:
(C) : Given,
\(\int \frac{3 \mathrm{x}+1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)} \mathrm{dx}=\mathrm{A} \log |\mathrm{x}-1|+\mathrm{B} \log |\mathrm{x}-2|+\) \(\mathrm{Clog}|\mathrm{x}-3|+\mathrm{c}\)
L.H.S., \(I=\int \frac{3 x+1}{(x-1)(x-2)(x-3)} \cdot d x\)
\(\because \frac{3 \mathrm{x}+1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}-2}+\frac{\mathrm{C}}{\mathrm{x}-3}\)
\(3 \mathrm{x}+1=\mathrm{A}(\mathrm{x}-2)(\mathrm{x}-3)+\mathrm{B}(\mathrm{x}-1)(\mathrm{x}-3)+\mathrm{C}(\mathrm{x}-1)\) \((\mathrm{x}-2)\)
For \(\mathrm{x}=1\),
\(3+1=\mathrm{A}(1-2)(1-3)+\mathrm{B}(1-1)(1-3)(1-3)+\)
\(4=2 \mathrm{~A} \Rightarrow \mathrm{A}=2\)
\(C(1-1)(1-2)\)
For, \(\mathrm{x}=2\),
\(7=\mathrm{A} \times 0-\mathrm{B}+\mathrm{C} \times 0\)
\(\mathrm{~B}=-7\)
For \(\mathrm{x}=3\),
\(10=A \times 0+B \times 0+2 C\)
\(C=5\)
\(\therefore \quad \int \frac{3 x+1}{(x-1)(x-2)(x-3)}=\int\left[\frac{2}{x-1}+\frac{(-7)}{x-2}+\frac{5}{x-3}\right] d x\)
\(=2 \log |\mathrm{x}-1|-7 \log |\mathrm{x}-2|+5 \log |\mathrm{x}-3|+\mathrm{c}\)
On comparing L.H.S. and R.H.S., we get -
\(\mathrm{A}=2, \mathrm{~B}=-7, \mathrm{C}=5\)