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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{x\sqrt{2x+3}}\ dx=\]     [AISSE 1985]

    A) \[\frac{x}{3}{{(2x+3)}^{3/2}}-\frac{1}{15}{{(2x+3)}^{5/2}}+c\]

    B) \[\frac{x}{3}{{(2x+3)}^{3/2}}+\frac{1}{15}{{(2x+3)}^{5/2}}+c\]

    C) \[\frac{x}{2}{{(2x+3)}^{3/2}}+\frac{1}{6}{{(2x+3)}^{5/2}}+c\]

    D) None of these

    Correct Answer: A

    Solution :

    • \[\int_{{}}^{{}}{x{{(2x+3)}^{1/2}}dx}\]                              
    • \[=x\frac{{{(2x+3)}^{3/2}}}{3/2}\frac{1}{2}-\int_{{}}^{{}}{\frac{{{(2x+3)}^{3/2}}}{3/2}\frac{1}{2}\,dx+c}\]                                
    • \[=\frac{1}{3}x{{(2x+3)}^{3/2}}-\frac{1}{3}\int_{{}}^{{}}{{{(2x+3)}^{3/2}}dx+c}\]                                           
    • \[=\frac{1}{3}x{{(2x+3)}^{3/2}}-\frac{1}{15}{{(2x+3)}^{5/2}}+c.\]


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