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

JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{(1+{{x}^{2}})\sqrt{{{p}^{2}}+{{q}^{2}}{{({{\tan }^{-1}}x)}^{2}}}}}=\]

    A) \[\frac{1}{q}\log [q{{\tan }^{-1}}x+\sqrt{{{p}^{2}}+{{q}^{2}}{{({{\tan }^{-1}}x)}^{2}}}]+c\]

    B) \[\log [q{{\tan }^{-1}}x+\sqrt{{{p}^{2}}+{{q}^{2}}{{({{\tan }^{-1}}x)}^{2}}}]+c\]

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

    D) None of these

    Correct Answer: A

    Solution :

    • Putting \[q{{\tan }^{-1}}x=t\]  \[\Rightarrow \frac{q}{1+{{x}^{2}}}dx=dt\Rightarrow \frac{1}{1+{{x}^{2}}}dx=\frac{dt}{q}\]                   
    • \[\Rightarrow \int_{{}}^{{}}{\frac{dx}{(1+{{x}^{2}})\sqrt{{{p}^{2}}+{{q}^{2}}{{({{\tan }^{-1}}x)}^{2}}}}}=\frac{1}{q}\int_{{}}^{{}}{\frac{dt}{\sqrt{{{p}^{2}}+{{t}^{2}}}}}\]                
    • \[=\frac{1}{q}\log \left[ q{{\tan }^{-1}}x+\sqrt{{{p}^{2}}+{{q}^{2}}{{({{\tan }^{-1}}x)}^{2}}} \right]+c\].


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