Question

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Answer 1

Ответ:

$1)\ \ y=cosec(4x-1)=\dfrac{1}{sin(4x-1)}\ \ ,\ \ \ \ \Big(\dfrac{C}{v}\Big)'=-\dfrac{C\, v'}{v^2} \\\\\\y'=-\dfrac{4cos(4x-1)}{sin^2(4x-1)}\\\\\\2)\ \ y=\dfrac{x^2}{1+cos4x}\\\\\\y'=\dfrac{2x\, (1+cos4x)-x^2\cdot (-4sin4x)}{(1+cos4x)^2}=\dfrac{2x+2x\, cos4x+4x^2\, sin4x}{(1+cos4x)^2}$

$3)\ \ \displaystyle \int \frac{dx}{1+2x^2}=\int \frac{dx}{1+(\sqrt2x)^2}=\frac{1}{\sqrt2}\, arctg(\sqrt2x)+C\\\\\\4\ \ \int \frac{2x\, dx}{\sqrt{3x}}=\frac{2}{\sqrt3}\int \frac{x\, dx}{\sqrt{x}}=\frac{2}{\sqrt3}\int \sqrt{x}\, dx=\frac{2}{\sqrt3}\cdot \frac{2\sqrt{x^3}}{3}+C=\frac{4\sqrt{x^3}}{3\sqrt3}+C$