Questions in definite-integral

Select	Question
	$\int_{\,0}^{\,1}{\frac{d}{dx}\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]\,dx}$ is equal to
	Let $f(x)=\int_{\,1}^{\,x}{\sqrt{2-{{t}^{2}}}dt}$. Then real roots of the equation ${{x}^{2}}-{f}'(x)=0$ are
	$\int_{\,0}^{\,\infty }{\frac{xdx}{(1+x)(1+{{x}^{2}})}=}$
	$\int_{0}^{a}{{{x}^{4}}\sqrt{{{a}^{2}}-{{x}^{2}}}}\,dx=$
	$\int_{0}^{a}{x{{(2ax-{{x}^{2}})}^{\frac{3}{2}}}\,dx=}$
	$\int_{0}^{a}{{{x}^{2}}{{({{a}^{2}}-{{x}^{2}})}^{3/2}}dx=}$
	Let$\frac{d}{dx}F(x)=\left( \frac{{{e}^{\sin x}}}{x} \right)\,;\,x>0$. If $\int_{\,1}^{\,4}{\frac{3}{x}{{e}^{\sin {{x}^{3}}}}dx=F(k)-F(1)}$, then one of the possible value of k, is
	If $f(x)=\int_{0}^{x}{t\sin t\,dt\,,}$ then ${f}'(x)=$
	$\underset{n\to \infty }{\mathop{\text{lim}\,}}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}}+.....+\frac{1}{n}{{\sec }^{2}}1 \right]$ equals
	Area bounded by the curve $y=\log x\,,$ $x-$axis and the ordinates $x=1,\,\,x=2$ is