JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{5}}\ dx}{\sqrt{(1+{{x}^{3}})}}=}\]        [IIT 1975]

    A)            \[\frac{2}{3}\sqrt{(1+{{x}^{3}})}({{x}^{3}}+2)\]

    B)            \[\frac{2}{9}\sqrt{(1+{{x}^{3}})}({{x}^{3}}-4)\]

    C)            \[\frac{2}{9}\sqrt{(1+{{x}^{3}})}({{x}^{3}}+4)\]

    D)            \[\frac{2}{9}\sqrt{(1+{{x}^{3}})}({{x}^{3}}-2)\]

    Correct Answer: D

    Solution :

                       Here \[{{x}^{5}}={{x}^{3}}{{x}^{2}}\] and differential coefficient of \[{{x}^{3}}\] is \[3{{x}^{2.}}\] In order to remove fractional powers, we put            \[1+{{x}^{3}}={{t}^{2}}\Rightarrow 3{{x}^{2}}dx=2t.\,dt\,;\] Also \[{{x}^{3}}={{t}^{2}}-1\]            \[I=\int_{{}}^{{}}{\frac{({{t}^{2}}-1)}{t}\left( \frac{2}{3}t\,dt \right)=\frac{2}{3}\int_{{}}^{{}}{({{t}^{2}}-1)\,dt}}\]              \[=\frac{2}{3}\left( \frac{{{t}^{3}}}{3}-t \right)=\frac{2}{9}t\,({{t}^{2}}-3)\]= \[\frac{2}{9}\sqrt{(1+{{x}^{3}})}\,({{x}^{3}}-2)\]


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