Question

Solve: $\int \frac{3x-1}{(x-1)(x-2)(x-3)}$

Answer 1

By partial fractions

$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{c}{x-3}$

$\therefore (3x-1)=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Put $x=1$

$\therefore 2=A(-1)(-2)$

$\therefore A=1$

Put $x=2$

$\therefore 5=B\left(1\right)(-1)$

$B=-5$

Put $x=3$

$\therefore 8=C\left(2\right)\left(1\right)$

$C=4$

$\therefore \int \frac{(3x-1)dx}{(x-1)(x-2)(x-3)}=\int \frac{dx}{x-1}+\int \frac{-5dx}{x-2}+\int \frac{4dx}{x-3}$

$=log|x-1|-5log|x-2|+4log(x-3)+c$

$=log(x-1)-log(x-2{)}^5+log(x-3{)}^4+c$

$=log\left[\frac{(x-1)(x-3{)}^4}{(x-2{)}^5}\right]+c$.