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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} $

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  • 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)}=$$\log |x+a|+c$
Step 1:
we can resolve this into parital fractions
Comparing the coefficient of like terms:
equating the coefficient of $x^2$
equating the coefficient of $x$
equating the constant terms
eliminating C from equ(1) and equ (2)
elimianting C from equ (2) and equ(3)
Multiply equ(4) by +4
$\qquad B=\large\frac{-1}{3}$
Substituting for B in equ (4)
$2A+5 (\large-\frac{1}{3})=$$-2$
$2A-\large\frac{5}{3}$$=-2 =>2A=-2+\large\frac{5}{3}$
Hence $A=\large\frac{-1}{6};$$\;B=\large\frac{-1}{3}\;$$ and \;C=\large\frac{1}{2}$
Now substituting the values
Step 2:
Now integrating we get,
$I=\large-\frac{1}{6} \int \frac{1}{x-1} dx -\frac{1}{3} \int \frac{dx}{x+2}+\frac{1}{2} \int \frac{dx}{x-3}$
$=\large\frac{-1}{6} $$\log |x-1|-\large\frac{1}{3} $$\log |x+2|+\large\frac{1}{2} $$\log |x-3|+c$
$=\large\frac{\log \sqrt {x-3}}{\log(x-1)^{1/6} +\log (x+2)^{1/3}}+c$
$=\log\bigg|\large\frac{\log \sqrt {x-3}}{\log(x-1)^{1/6} \times \log (x+2)^{1/3}}\bigg|+c$
answered Apr 22, 2013 by meena.p
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