JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If \[\int_{{}}^{{}}{\frac{dx}{{{\left( {{x}^{2}}-2x+10 \right)}^{2}}}}\]\[=A\left( {{\tan }^{-1}}\left( \frac{x-1}{3} \right)+\frac{f(x)}{{{x}^{2}}-2x+10} \right)+C\] where C is a constant of integration, then : [JEE Main 10-4-2019 Morning]

    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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