If \(\int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x A \log |x-1| B\) \(\log |x-2|+C \log |x-3|+C\), then the values of \(A, B\) and \(C\) are respectively

We have,
\(\int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x\)
Let \(\frac{3 x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\)
\(\Rightarrow \quad(3 x+1)=A(x-2)(x-3)+B(x-1)(x-3)\)
\(+C(x-1)(x-2)\)
Put \(x-1=0\)
\(\Rightarrow \quad x=1\)
Then, \(3 \times 1+1=A(-1)(-2)\)
\(\Rightarrow \quad 4=2 A\)
\(\Rightarrow \quad A=2\)
Put \(\quad x-2=0\)
\(\Rightarrow \quad x=2\)
Then, \(7=B(2-1)(2-3)\)
\(\Rightarrow \quad 7=B(1)(-1)\)
\(\Rightarrow \quad B=-7\)
And put \(x-3=0\)
\(\Rightarrow \quad x=3\)
Then, \(10=C(3-1)(3-2)\)
\(\Rightarrow \quad 10=C(2)(1)\)
\(\Rightarrow \quad C=5\)
\(\therefore \quad \frac{3 x+1}{(x-1)(x-2)(x-3)}=\frac{2}{x-1}-\frac{7}{x-2}+\frac{5}{x-3}\)
\(\therefore \quad \int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x\)
\(\quad=\int \frac{2}{x-1} d x-\int \frac{7}{x-2} d x+\int \frac{5}{x-3} d x\)
\(\Rightarrow \quad \int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x\)
\(=2 \log |x-1|-7 \log |x-2|+5 \log |x-3|+C\)
\(=A \log |x-1|+B \log |x-2|+C \log |x-3|+C\)
\((\) Given \()\)
\(\Rightarrow A=2, B=-7, C=5\)