Definite Integration Question 406
Question: $ \int_0^{1}{\frac{dx}{\sqrt{1+x}-\sqrt{x}}=} $
[SCRA 1986]
Options:
A) $ \frac{2\sqrt{2}}{3} $
B) $ \frac{4\sqrt{2}}{3} $
C) $ \frac{8\sqrt{2}}{3} $
D) None of these
Show Answer
Answer:
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} $ .
BETA
