Integral Calculus Question 494
Question: $ \int{\sqrt{\frac{1+x}{1-x}}dx=} $
[RPET 2002]
Options:
A) $ -{{\sin }^{-1}}x-\sqrt{1-x^{2}},+c $
B) $ {{\sin }^{-1}}x+\sqrt{1-x^{2}},+c $
C) $ {{\sin }^{-1}}x-\sqrt{1-x^{2}},+c $
D) $ -{{\sin }^{-1}}x-\sqrt{x^{2}-1},+c $
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Answer:
Correct Answer: C
Solution:
$ I=\int{\sqrt{\frac{1+x}{1-x}}dx} $ $ =\int{\frac{1+x}{\sqrt{1-x^{2}}}dx} $ $ =\int{\frac{dx}{\sqrt{1-x^{2}}}+\int{\frac{x}{\sqrt{1-x^{2}}}dx}} $ $ ={{\sin }^{-1}}x-\sqrt{1-x^{2}}+c $ .
BETA
