Questions in definite-integral

Select	Question
	$\int_{0}^{1}{\sqrt{\frac{1-x}{1+x}}}\,dx$ equals
	$\int_{1}^{e}{\frac{1}{x}\,dx}$ is equals to
	$\int_{\,1}^{\,x}{\frac{\log {{x}^{2}}}{x}\,dx=}$
	$\frac{1}{2}(e-3)$
	$\int_{0}^{\pi /4}{{}}(\cos x-\sin x)dx+\int_{\pi /4}^{5\pi /4}{{}}(\sin x-\cos x)dx + \int_{2\pi }^{\pi /4}{{}}(\cos x-\sin x)\,dx$ is equal to
	$\left( \int_{\,0}^{\,a}{x\,dx} \right)\le (a+4),$ then
	The value of $\int_{\,-\,1}^{\,3}{\,{{\tan }^{-1}}\left( \frac{x}{{{x}^{2}}+1} \right)+{{\tan }^{-1}}\left( \frac{{{x}^{2}}+1}{x} \right)\,dx}$ is
	$\int_{\,-\,1}^{\,0}{\frac{dx}{{{x}^{2}}+2x+2}=}$
	The value of $\int_{\,0}^{\,1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}$ is
	$\int_{1}^{\sqrt{3}}{\frac{1}{1+{{x}^{2}}}dx}$ is equal to