Question

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

Answer 1

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

Let,

$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}....\left(1\right)$

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

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

$3x-1=(A+B+C)x^2-(5A+4B+3C)x+(6A+3B+2C)$

Now, comparing coefficients of all powers of $x$ both sides of above equation we get

$A+B+C=0...........\left(2\right)$

$-(5A+4B+3C)=3....\left(3\right)$

$6A+3B+2C=-1......\left(4\right)$

Solving the above system of equation we get:

$A=1;B=-5;C=4;$

Put values of $A,B,andC$ in equation (1) we get:

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

Hence,

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

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

or

$\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx=\ln (x-1)-\ln \left(\right(x-2{)}^5)+\ln ((x-3{)}^4)\because a\ln \left(x\right)=\ln \left(x^a\right)$

$\int \frac{3x-1}{(x-1)(x-2)(x-3)}dx=\ln (\frac{(x-1)(x-3{)}^4}{(x-2{)}^5})\because \ln \left(a\right)+\ln \left(b\right)=\ln \left(ab\right)and\ln \left(a\right)-\ln \left(b\right)=\ln (\frac{a}{b})$