Question

Integrate the rational functions.
$\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx.$

Answer 1

Let $\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$
$\Rightarrow \frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\Rightarrow 3x-1=A[x^2-5x+6]+B[x^2-4x+3]+C[x^2-3x+2]\Rightarrow 3x-1=x^2(A+B+C)+x(-5A-4B-3C)+(6A+3B+2C)$

On equating the coefficients of $x^2$, x and constant term on both sides, we get
A+B+C =0 ......(i)
-5A-4B-3C =3 .......(ii)
and 6A+3B+2C=-1 .....(iii)
Form Eq.(i), we get A =-(B+C)
On putting the value of A in Eqs. (ii) and (iii), we get
$-5(-(B+C\left)\right)-4B-3C=3\Rightarrow 5B+5C-4B-3C=3$
$\Rightarrow B+2C=3......\left(iv\right)$
and $6(-(B+C\left)\right)+3B+2C=-1\Rightarrow -6B-6C+3B+2C=-1$
$\Rightarrow -3B-4C=-1..........\left(v\right)$
On solving Eqs. (iv)and (v), we get $\Rightarrow C=4$
On putting the value of C in Eq. (iv), we get $B+2\times 4=3\Rightarrow B=-5$
Putting the value of B and C in Eq. (i), we get A +(-5)+4 =0 $\Rightarrow A=1\therefore A=1,B=-5,C=4$
Now, $\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx=\int (\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)})dx$
$=\int \frac{1}{(x-1)}dx+\int \frac{(-5)}{(x-2)}dx+\int \frac{4}{(x-3)}dx=log|x-1|-5log|x-2|+4log|x-3|+C(\because \frac{1}{x}dx=logx)$