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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{5}}}{\sqrt{1+{{x}^{3}}}}dx=}\]      [IIT 1985]

    A) \[\frac{2}{9}{{(1+{{x}^{3}})}^{3/2}}+c\]

    B) \[\frac{2}{9}{{(1+{{x}^{3}})}^{3/2}}+\frac{2}{3}{{(1+{{x}^{3}})}^{1/2}}+c\]

    C) \[\frac{2}{9}{{(1+{{x}^{3}})}^{3/2}}-\frac{2}{3}{{(1+{{x}^{3}})}^{1/2}}+c\]

    D) None of these

    Correct Answer: C

    Solution :

    • Put \[1+{{x}^{3}}={{t}^{2}}\Rightarrow 3{{x}^{2}}dx=2t\,dt\] and \[{{x}^{3}}={{t}^{2}}-1\]                   
    • So, \[\int_{{}}^{{}}{\frac{{{x}^{5}}}{\sqrt{1+{{x}^{3}}}}\,dx=\int_{{}}^{{}}{\frac{{{x}^{2}}.{{x}^{3}}}{\sqrt{1+{{x}^{3}}}}\,dx}}\]                   
    • \[=\frac{2}{3}\int_{{}}^{{}}{\frac{({{t}^{2}}-1)\,.\,t\,dt}{t}=\frac{2}{3}\int_{{}}^{{}}{({{t}^{2}}-1)\,dt=\frac{2}{3}\left[ \frac{{{t}^{3}}}{3}-t \right]}}+c\]
    • \[=\frac{2}{3}\left[ \frac{{{(1+{{x}^{3}})}^{3/2}}}{3}-{{(1+{{x}^{3}})}^{1/2}} \right]+c\].


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