If $$\mathrm{f}(x)=\frac{\mathrm{e}^x}{1+\mathrm{e}^x}, \mathrm{I}_1=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} x \mathrm{~g}(x(1-x)) \mathrm{d} x$$ and $$\mathrm{I}_2=\int_\limits{\mathrm{f}(-\mathrm{a})}^{\mathrm{f}(\mathrm{a})} \mathrm{g}(x(1-x)) \mathrm{d} x$$, then the value of $$\frac{I_2}{I_1}$$ is
Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a differentiable function and $$f(1)=4$$. Then the value of $$\lim _\limits{x \rightarrow 1} \int_\limits4^{f(x)} \frac{2 t}{x-1} d t$$, if $$f^{\prime}(1)=2$$ is
$$ \text { If } \int \frac{\log _e\left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\mathrm{f}(\mathrm{g}(x))+\mathrm{c} \text { then } $$
Let $$\mathrm{I}(\mathrm{R})=\int_\limits0^{\mathrm{R}} \mathrm{e}^{-\mathrm{R} \sin x} \mathrm{~d} x, \mathrm{R}>0$$. then,