A) \[-{{\sin }^{-1}}x\sqrt{1-{{x}^{2}}}+c\]
B) \[{{\sin }^{-1}}x-\sqrt{{{x}^{2}}-1}+c\]
C) \[{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}+c\]
D) \[-{{\sin }^{-1}}x-\sqrt{{{x}^{2}}-1}+c\]
Correct Answer: C
Solution :
Let \[I=\int{\sqrt{\frac{1+x}{1-x}}}dx\] \[=\int{\frac{1+x}{\sqrt{1-{{x}^{2}}}}}dx\] \[=\int{\frac{1}{\sqrt{1-{{x}^{2}}}}}dx+\int{\frac{x}{\sqrt{1-{{x}^{2}}}}}dx\] Let \[1-{{x}^{2}}=t\Rightarrow -2xdx=dt\] \[\Rightarrow \] \[xdx=-\frac{1}{2}dt\] \[\therefore \] \[I={{\sin }^{-1}}x+\int{\frac{1}{\sqrt{t}}\left( -\frac{1}{2} \right)}dt\] \[={{\sin }^{-1}}x-\frac{1}{2}\left( \frac{{{t}^{-\frac{1}{2}+1}}}{-\frac{1}{2}+1} \right)+e\] \[={{\sin }^{-1}}x-\sqrt{t}+c\] \[={{\sin }^{-1}}x-\sqrt{1+{{x}^{2}}}+c\]
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