JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int {\frac{{dx}}{{(1 + \sqrt x ) \cdot \sqrt x \sqrt {1 - x} }}} \) is equal to (where \(c\) is a constant of integration)
- A \( - 2\sqrt {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}} + c\)
- B \( - \sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} + c\)
- C \( - 2\sqrt {\frac{{1 - \sqrt x }}{{1 +\sqrt x }}} + c\)
- D \( 2\sqrt {\frac{{1 + \sqrt x }}{{1 - \sqrt x }}} + c\)
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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