$$ \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} d x= $$

Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. 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 $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$

The differential equation of the family of circles with fixed radius $r$ units and centre on the line $y=3$, is

The degree of the differential equation

$$ x\left(\frac{d^2 y}{d x^2}\right)^{1 / 3}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{5 / 3}+7 \frac{d y}{d x}+y=0 $$