1.

\[\int_{{}}^{{}}{\frac{1+{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx=}\]    [IIT 1977]

A.            \[\frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
B.            \[\frac{3}{2}{{\sin }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
C.            \[\frac{3}{2}{{\cos }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
D.            \[\frac{3}{2}{{\cos }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]
Answer» B.            \[\frac{3}{2}{{\sin }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c\]


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