$\int_{0}^{100} e^{x-[x]} d x$ is equal to

Let $I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x}\left(\frac{3}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$. Then $3 I^2=$

$\int_{-\pi / 15}^{\pi / 15} \frac{\cos 5 x}{1+e^{5 x}} d x=$

If $f(x) = \int\limits_1^x \frac{\ln t}{1 + t} dt$ where $x > 0$,then the value$(s)$ of $x$ satisfying the equation $f(x) + f(1/x) = 0$ is/are:

Evaluate $\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$

The value of the integral $\int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \frac{1}{x} \right) dx$ is