# Evaluate the following: $\large\int\frac{2x-1}{(x-1)(x+2)(x-3)}dx$

$\begin{array}{1 1} (A)\;\log\bigg|\large\frac{\log \sqrt {x+3}}{\log(x-1)^{1/6} \times \log (x+2)^{1/3}}\bigg| \\(B)\;\log\bigg|\large\frac{\log x-3}{\log(x-1)^{1/6} \times \log (x+2)^{1/3}}\bigg|+c \\ (C)\;\log\bigg|\large\frac{\log \sqrt {x-3}}{\log(x-1)^{1/6} \times \log (x+2)^{1/3}}\bigg|+c \\(D)\;\log\bigg|\large\frac{\log \sqrt {x-3}}{\log(x+1)^{1/6} \times \log (x-2)^{1/3}}\bigg|+c \end{array}$

Toolbox:
• A rational expression of the form $\large\frac{ax+b}{(x+c)(x+d)}$ can be resolved into partial fractions as $\large\frac{A}{(x+c)}+\frac{B}{(x+d)}$
• $\large\frac{dx}{(x+a)}=$$\log |x+a|+c Step 1: \large\int\frac{2x-1}{(x-1)(x+2)(x-3)}dx we can resolve this into parital fractions \large\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x-3)} =>2x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2) Comparing the coefficient of like terms: equating the coefficient of x^2 0=A+B+C-----(1) equating the coefficient of x 2=-A-4B+C------(2) equating the constant terms -1=-6A+3B-2C-----(3) eliminating C from equ(1) and equ (2) A+B+C=0 -A-4B+C=2 __________________ 2A+5B=-2------(4) elimianting C from equ (2) and equ(3) -A-4B+C=2 -2A-8B+2C=4 -6A+3B-2C=-1 ______________________ -8A-5B=3------(5) Multiply equ(4) by +4 +8A+20B=-8 -8A-5B=3 _______________________ \qquad15B=-5 \qquad B=\large\frac{-1}{3} Substituting for B in equ (4) 2A+5 (\large-\frac{1}{3})=$$-2$
$2A-\large\frac{5}{3}$$=-2 =>2A=-2+\large\frac{5}{3} 2A=-\large\frac{1}{3}=>$$A=-\large\frac{1}{6}$
$A+B+C=0$
$\large-\frac{1}{6}-\frac{1}{3}+$$C=0 C=\large\frac{1}{2} Hence A=\large\frac{-1}{6};$$\;B=\large\frac{-1}{3}\;$$and \;C=\large\frac{1}{2} Now substituting the values \large\frac{2x-1}{(x+1)(x+2)(x-3)}=\frac{-1/6}{(x-1)}+\frac{-1/3}{(x+2)}+\frac{1/2}{(x-3)} Step 2: Now integrating we get, I=\large-\frac{1}{6} \int \frac{1}{x-1} dx -\frac{1}{3} \int \frac{dx}{x+2}+\frac{1}{2} \int \frac{dx}{x-3} =\large\frac{-1}{6}$$\log |x-1|-\large\frac{1}{3} $$\log |x+2|+\large\frac{1}{2}$$\log |x-3|+c$
$=\large\frac{\log \sqrt {x-3}}{\log(x-1)^{1/6} +\log (x+2)^{1/3}}+c$
$=\log\bigg|\large\frac{\log \sqrt {x-3}}{\log(x-1)^{1/6} \times \log (x+2)^{1/3}}\bigg|+c$