$\int_{{}}^{{}}{\frac{1+{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx=}$

Question

Accepted Answer

$\frac{3}{2}{{\sin }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c$

Answer

$\frac{3}{2}{{\sin }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c$

Answer

$\frac{3}{2}{{\cos }^{-1}}x-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c$

Answer

$\frac{3}{2}{{\cos }^{-1}}x+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+c$

indefinite-integration