Questions in indefinite-integration

Select	Question
	$\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\ }dx=$
	$\int_{{}}^{{}}{\frac{{{\sin }^{-1}}x}{{{(1-{{x}^{2}})}^{3/2}}}\ dx=}$
	$\int_{{}}^{{}}{\frac{x{{\tan }^{-1}}x}{{{(1+{{x}^{2}})}^{3/2}}}\ dx=}$
	$\int_{{}}^{{}}{{{x}^{5}}.{{e}^{{{x}^{2}}}}dx=}$
	$\int_{{}}^{{}}{{{e}^{{{\tan }^{-1}}x}}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)\ dx$ is equal to
	$\int_{{}}^{{}}{{{e}^{\sqrt{x}}}\ dx}$ is equal to (A is an arbitrary constant)
	${{I}_{1}}=\int{{{\sin }^{-1}}x\,\,dx}$ and ${{I}_{2}}=\int{{{\sin }^{-1}}\sqrt{1-{{x}^{2}}}}dx$then
	If an antiderivative of $f(x)$ is ${{e}^{x}}$ and that of $g(x)$ is $\cos x,$then $\int{f(x)\cos x\,dx}+\int{g(x){{e}^{x}}dx=}$
	$\int_{{}}^{{}}{\frac{dx}{(x-{{x}^{2}})}=}$
	$\int_{{}}^{{}}{\frac{dx}{1+x+{{x}^{2}}+{{x}^{3}}}=}$