$\int \frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\mathrm{dx}$ is equal to

detailed solution

Correct option is A

Let 3x−1(x−1)(x−2)(x−3)=A(x−1)+B(x−2)+C(x−3)⇒3x−1(x−1)(x−2)(x−3)=A(x−2)(x−3)+B(x−1)(x−3)+C(x−1)(x−2)(x−1)(x−2)(x−3)⇒3x−1=Ax2−5x+6+Bx2−4x+3+Cx2−3x+2⇒3x−1=x2(A+B+C)+x(−5A−4B−3C)+(6A+3B+2C)On equating the coefficients of x2,x and constant term on both sides, we getA+B+C=0                  ......(i)−5A−4B−3C=3         .....(ii)           and 6A+3B+2C=−1      .....(iii)From Eq. (i), we get /, = -(B + C)On putting the value of A in Eqs. (ii) and (iii), we get     −5{−(B+C)}−4B−3C=3⇒    5B+5C−4B−3C=3⇒    B+2C=3                   .....(iv)and 6{−(B+C)}+3B+2C=−1⇒    −6B−6C+3B+2C=−1⇒    −3B−4C=−1                      ......(iv)On solving Eqs. (iv1 and (v), we get C = 4 On putting the value of C in Eq. (iv), we getB+2×4=3⇒B=−5On putting the value of Band C in Eq. (i), we getA+(−5)+4=0⇒A=1A=1,B=−5,C=4Now,∫3x−1(x−1)(x−2)(x−3)dx=∫A(x−1)+B(x−2)+C(x−3)dx=∫1(x−1)dx+∫(−5)(x−2)dx+∫4(x−3)dx=log⁡|x−1|−5|log⁡|x−2|+4log⁡|x−3∣+C ∵1xdx=log⁡x