2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Indefinite Integrals

JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\]            [IIT 1979]

    A) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]

    B) \[\frac{1}{{{b}^{2}}}\left[ x-\frac{2a}{b}\log (a+bx)+\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]

    C) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{2a}{b}\log (a+bx)+\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]

    D) \[\frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{a+bx} \right]\]

    Correct Answer: D

    Solution :

    • Put \[a+bx=t\Rightarrow x=\frac{t-a}{b}\] and \[dx=\frac{dt}{b}\]                   
    • \[\therefore \,\,\,I={{\int_{{}}^{{}}{\left( \frac{t-a}{b} \right)}}^{2}}\times \frac{1}{{{t}^{2}}}\frac{dt}{b}\]                   
    • \[=\frac{1}{{{b}^{2}}}\int_{{}}^{{}}{\left( 1-\frac{2a}{t}+{{a}^{2}}.{{t}^{-2}} \right)}\,dt=\frac{1}{{{b}^{2}}}\left[ t-2a\,\,\log t-\frac{{{a}^{2}}}{t} \right]\]
    • \[=\frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{(a+bx)} \right]\].


You need to login to perform this action.
You will be redirected in 3 sec spinner