A) \[A=\frac{1}{27}\text{and}\,f(x)=9(x-1)\]
B) \[A=\frac{1}{81}\text{and}\,f(x)=3(x-1)\]
C) \[A=\frac{1}{54}\text{and}\,f(x)=9{{(x-1)}^{2}}\]
D) \[A=\frac{1}{54}\text{and}\,f(x)=3(x-1)\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{dx}{{{\left( {{\left( x-1 \right)}^{2}}+9 \right)}^{2}}}}=\frac{1}{27}\int_{{}}^{{}}{{{\cos }^{2}}\theta d\theta }\] \[(Put\,x-1=3tan\theta )\] \[=\frac{1}{54}\int_{{}}^{{}}{\left( 1+\cos 2\theta \right)}d\theta =\frac{1}{54}\left( \theta +\frac{\sin 2\theta }{2} \right)+C\] \[=\frac{1}{54}\left( {{\tan }^{-1}}\left( \frac{x-1}{3} \right)+\frac{3\left( x-1 \right)}{{{x}^{2}}-2x+10} \right)+C\]
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