A) \[\frac{2\sqrt{2}}{3}\]
B) \[\frac{4\sqrt{2}}{3}\]
C) \[\frac{8\sqrt{2}}{3}\]
D) None of these
Correct Answer: B
Solution :
\[I=\int_{0}^{1}{\frac{dx}{\sqrt{1+x}-\sqrt{x}}=\int_{0}^{1}{\frac{(\sqrt{1+x}+\sqrt{x})dx}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}}}\] \[=\int_{0}^{1}{\frac{(\sqrt{1+x}+\sqrt{x})}{1+x-x}}dx=\int_{0}^{1}{\sqrt{1+x\,}dx}+\int_{0}^{1}{\sqrt{x}}dx=\frac{4\sqrt{2}}{3}\].
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