Question:
$\frac{3 x-1}{(x-1)(x-2)(x-3)}$
Solution:
Let $\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$
$3 x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ ...(1)
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain
A = 1, B = −5, and C = 4
$\therefore \frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}$
$\Rightarrow \int \frac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int\left\{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\right\} d x$
$=\log |x-1|-5 \log |x-2|+4 \log |x-3|+C$
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