Question

Integrate the rational function: $\frac{x}{(x-1)(x-2)(x-3)}$

Answer 1

$\int \frac{x}{(x-1)(x-2)(x-3)}.dx$
Using partial fraction $\Rightarrow \frac{x}{(x-1)(x-2)(x-3)}$
$=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}$
$\Rightarrow x=A(x-2)(x-3)$
$+B(x-1)(x-3)$
$+C(x-1)(x-2)$
Putting x =1
$\Rightarrow 1=A(1-2)(1-3)$
$+B(1-1)(1-3)$
$+C(1-1)(1-2)$
$\Rightarrow 1=A(-1)(-2)$
$\Rightarrow A=\frac{1}{2}$
Similarly putting x =2,
$\Rightarrow 2$
$=A(2-2)(2-3)+B(2-1)(2-3)+C(2-1)(2-2)$
$\Rightarrow 2=B\left(1\right)(-1)$
$\Rightarrow B=-2$
Similarly putting x =3,
$\Rightarrow 3=A(3-2)(3-3)$
$+B(3-1)(3-3)$
$+C(3-1)(3-2)$
$\Rightarrow 3=C\times \left(2\right)\left(1\right)$
$\Rightarrow C=\frac{3}{2}$
Now, putting the values of A, B and C
$\Rightarrow \frac{x}{(x-1)(x-2)(x-3)}$
$=\frac{\frac{1}{2}}{x-1}+\frac{-2}{x-2}+\frac{\frac{3}{2}}{x-3}$
$=\frac{1}{2(x-1)}-\frac{2}{x-2}+\frac{3}{2(x-3)}$
Now,
$\int \frac{x}{(x-1)(x-2)(x-3)}dx$
$=\int (\frac{1}{2(x-1)}-\frac{2}{x-2}+\frac{3}{2(x-3)}).dx$
$=\frac{1}{2}\int \frac{1}{x-1}.dx-2\int \frac{1}{x-2}.dx$
$+\frac{3}{2}\int \frac{1}{x-3}.dx$
$=\frac{1}{2}\log |x-1|-2\log |x-2|$
$+\frac{3}{2}\log |x-3|+C$
Where C is constant of integration