JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Fundamental Integration

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{\sqrt{x}+\sqrt{x-2}}=}\] [MP PET 1990]

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

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

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

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

    Correct Answer: A

    Solution :

               \[\int_{{}}^{{}}{\frac{dx}{\sqrt{x}+\sqrt{x-2}}=\frac{1}{2}\int_{{}}^{{}}{\frac{x-(x-2)}{\sqrt{x}+\sqrt{x-2}}\,dx}}\]                   \[=\frac{1}{2}\int_{{}}^{{}}{(\sqrt{x}-\sqrt{x-2})\,dx}=\frac{1}{2}\left[ \frac{{{x}^{3/2}}}{3/2}-\frac{{{(x-2)}^{3/2}}}{3/2} \right]+c\]                   \[=\frac{1}{3}\left\{ {{x}^{3/2}}-{{(x-2)}^{3/2}} \right\}+c.\]


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