A) \[\frac{11}{12},\frac{5}{6}\]
B) \[\frac{11}{12},\frac{-5}{6}\]
C) \[-\frac{11}{12},\frac{5}{6}\]
D) None of these
Correct Answer: A
Solution :
\[I=\int{\frac{2{{x}^{2}}+3}{({{x}^{2}}-1)({{x}^{2}}-4)}}\] \[\frac{2{{x}^{2}}+3}{({{x}^{2}}-1)({{x}^{2}}-4)}=\frac{A}{{{x}^{2}}-1}+\frac{B}{{{x}^{2}}-4}\] \[2{{x}^{2}}+3=A({{x}^{2}}-4)+B({{x}^{2}}-1)\] \[2{{x}^{2}}+3={{x}^{2}}(A+B)-4A-B\] Comparing the coefficient of \[{{x}^{2}}\]and constant term on both sides, \[A+B=2\] .....(i) \[4A+B=-3\] .....(ii) On solving both the equations \[A=-\frac{5}{3}\], \[B=\frac{11}{3}\] \[\int{\frac{2{{x}^{2}}+3.dx}{({{x}^{2}}-1)({{x}^{2}}-4)}}\]\[=\int{\frac{-\frac{5}{3}dx}{({{x}^{2}}-1)}+\int{\frac{\frac{11}{3}dx}{({{x}^{2}}-4)}}}\] \[=-\frac{5}{3}\int{\frac{dx}{(x+1)(x-1)}}\]\[+\frac{11}{3}\int{\frac{dx}{(x+2)(x-2)}}\] \[=-\frac{5}{3}.\frac{1}{2}\int{\frac{dx}{x-1}+\frac{5}{6}\int{\frac{dx}{x+1}}}+\frac{11}{3}.\frac{1}{4}\int{\frac{dx}{x-2}-\frac{11}{12}\int{\frac{dx}{x+2}+c}}\] \[=-\frac{5}{6}\log (x-1)+\frac{5}{6}\]log(x+1)\[+\frac{11}{12}\]log(x?2)\[-\frac{11}{12}\]log(x+2)+c \[=\frac{5}{6}\log \left( \frac{x+1}{x-1} \right)+\frac{11}{12}\log \left( \frac{x-2}{x+2} \right)+c\] \[=\log {{\left( \frac{x+1}{x-1} \right)}^{5/6}}+\log {{\left( \frac{x-2}{x+2} \right)}^{11/12}}+c\] Þ \[a=\frac{11}{12}\] and \[b=\frac{5}{6}\].
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