A) \[\frac{1}{3}\log \left| \frac{y-2}{y+1} \right|=\frac{1}{4}\log \left| \frac{x+3}{x-1} \right|+c\]
B) \[\frac{1}{3}\log \left| \frac{y+1}{y-2} \right|=\frac{1}{4}\log \left| \frac{x-1}{x+3} \right|+c\]
C) \[4\log \left| \frac{y-2}{y+1} \right|=3\log \left| \frac{x-1}{x+3} \right|+c\]
D) None of these
Correct Answer: C
Solution :
\[\frac{dy}{dx}=\frac{{{y}^{2}}-y-2}{{{x}^{2}}+2x-3}\] Þ \[\frac{dy}{(y-2)(y+1)}=\frac{dx}{(x+3)(x-1)}\] Þ \[\int_{{}}^{{}}{\frac{dy}{(y-2)(y+1)}}=\int_{{}}^{{}}{\frac{dx}{(x+3)(x-1)}}\] Þ \[\frac{1}{3}\int_{{}}^{{}}{\left( \frac{1}{y-2}-\frac{1}{y+1} \right)}dy=\frac{1}{4}\int_{{}}^{{}}{\left( \frac{1}{x-1}-\frac{1}{x+3} \right)\,}dx\] Þ \[\frac{1}{3}\log \left| \frac{y-2}{y+1} \right|=\frac{1}{4}\left| \frac{x-1}{x+3} \right|+c\].
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