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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

\(\int {\frac{{dx}}{{(1 + \sqrt x ) \cdot \sqrt x \sqrt {1 - x} }}} \) મેળવો.

  1. A \( - 2\sqrt {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}}  + c\)
  2. B \( - \sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}}  + c\)
  3. C \( - 2\sqrt {\frac{{1 - \sqrt x }}{{1 +\sqrt x }}}  + c\)
  4. D \( 2\sqrt {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}}  + c\)
Verified Solution

Answer & Solution

Correct Answer

(C) \( - 2\sqrt {\frac{{1 - \sqrt x }}{{1 +\sqrt x }}}  + c\)

Step-by-step Solution

Detailed explanation

\(I = \int {\frac{{dx}}{{(1 + \sqrt x ) \cdot \sqrt x \sqrt {1 - x} }}} \) \({\rm{ Put }}1 + \sqrt x = t\) \( \Rightarrow \frac{1}{{2\sqrt x }}dx = dt\) \(\Rightarrow \quad I=\int \frac{2 d t}{t \sqrt{2 t-t^{2}}}\) Again put \(t=\frac{1}{z}\)…
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