$\int \frac{3\mathrm{x}-1}{(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)}\mathrm{dx}$ 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,xand constant term on both sides, we getA+B+C=0----i−5A−4B−3C=3----ii6A+3B+2C=−1----iiiFrom Eq. (i), we get A = -(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----vOn solving Eqs. (iv1 and (v), we get C = 4On 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=1∴ A=1,B=−5,C=4 Now, ∫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|−5log⁡|x−2|+4log⁡|x−3|+C ∵1xdx=log⁡x