JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

  • question_answer
    If \[\int{\frac{dx}{{{x}^{22}}({{x}^{7}}-6)}}\]\[=A\{In{{(p)}^{6}}+9{{p}^{2}}-2{{p}^{3}}-18p\}+c\] then

    A) \[A=\frac{1}{9072},p=\left( \frac{{{x}^{7}}-6}{{{x}^{7}}} \right)\]

    B) \[A=\frac{1}{54432},p=\left( \frac{{{x}^{7}}-6}{{{x}^{7}}} \right)\]

    C) \[A=\frac{1}{54432},p=\left( \frac{{{x}^{7}}}{{{x}^{7}}-6} \right)\]

    D) \[A=\frac{1}{9072},p={{\left( \frac{{{x}^{7}}-6}{{{x}^{7}}} \right)}^{-1}}\]

    Correct Answer: B

    Solution :

    [b] Let \[I=\int{\frac{dx}{{{x}^{29}}\left( 1-\frac{6}{{{x}^{7}}} \right)}}\]
    Put \[1-\frac{6}{{{x}^{7}}}=p\Rightarrow \frac{42}{{{x}^{8}}}dx=dp\] and \[{{x}^{7}}=\frac{6}{1-p}\]
    \[\therefore I=\frac{1}{42}\int{\frac{{{(1-p)}^{3}}}{{{(6)}^{3}}p}dp}\]
    \[=\frac{1}{(42)(216)}\int{\frac{1-{{p}^{3}}-3p+3{{p}^{2}}}{p}dp}\]
    \[=\frac{1}{9072}\int{\left( \frac{1}{p}-{{p}^{2}}-3+3p \right)dp}\]
    \[=\frac{1}{9072}\left( \log p-\frac{{{p}^{3}}}{3}-3p+\frac{3}{2}{{p}^{2}} \right)+c\]
    \[=\frac{1}{54432}(6l\,np-2{{p}^{3}}-18p+9{{p}^{2}})+c\]
    \[=\frac{1}{54432}(l\,n{{p}^{6}}+9{{p}^{2}}-2{{p}^{3}}-18p)+c\]
    \[A=\frac{1}{54432},p=\left( \frac{{{x}^{7}}-6}{{{x}^{7}}} \right)\]


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