Question

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

Answer 1

$\int \frac{x}{(x-1)(x-2)(x-3)}dx$
Let $\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}$
$\Rightarrow \frac{x}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\Rightarrow x=A[x^2-2x-3x+6]+B[x^2-4x+3]+C[x^2-x-2x+2]\Rightarrow x=x^2(A+B+C)+x(-5A-4B-3C)+(6A+3B+2C)$
On comparing the coefficients of $x^2$, x and constant term on both sides, we get
$A+B+C=0\Rightarrow A=-(B+C)........\left(i\right)$
-5A -4B-3C=1 ...........(ii)
and 6A+3B +2C =0........(iii)
On putting the value of A in Eqs. (ii)and (iii), we get
$-5(-(B+C\left)\right)-4B-3C=1\Rightarrow 5B+5C-4B-3C=1\Rightarrow B+2C=1......\left(iv\right)$
and $6(-(B+C\left)\right)+3B+2C=0\Rightarrow -3B-4C=0........\left(v\right)$
Multiplying by 3 in Eq. (iv) and then adding in Eq. (v), we get C $=\frac{3}{2}$
On putting the value of C in Eq. (iv), we get $B+2\times \frac{3}{2}=1\Rightarrow B=-2.$
Now, put the values of B and C in Eq. (i), we get
$A-2+\frac{3}{2}=0\Rightarrow A=\frac{4-3}{2}=\frac{1}{2}\therefore A=\frac{1}{2},B=-2\text{and}C=\frac{3}{2}$
Now, $\int \frac{x}{(x-1)(x-2)(x-3)}dx=\int \frac{A}{(x-1)}dx+\int \frac{B}{(x-2)}dx+\int \frac{C}{(x-3)}dx$
$=\frac{1}{2}\int \frac{1}{(x-1)}dx-2\int \frac{1}{(x-2)}dx+\frac{3}{2}\int \frac{1}{(x-3)}dx=\frac{1}{2}log|x-1|-2log|x-2|+\frac{3}{2}log|x-3|+C$