tex]...

Daeir @Daeir

September 2021 1 46 Report

Интегральное исчисление функции! Помогите, пожалуйста!
[tex]\int\ {\frac{7xdx}{\sqrt{4-7x} } } \,[/tex]

Answers & Comments

Darknessich

Ответ:

$\frac{2(4-7x)\sqrt{4-7x} }{21} -\frac{8\sqrt{4-7x}}{7} +C$

Пошаговое объяснение:

$\int\limits {\frac{7x}{\sqrt{4-7x} } } \, dx =\int\limits {\frac{-(4-7x)+4}{\sqrt{4-7x} } } \, dx \\\\t = 4-7x,\ \frac{dt}{dx} = \frac{d(4-7x)}{dx} = \frac{d}{dx}(4) - \frac{d}{dx}(7x) = 0-7=-7\Rightarrow dx = \frac{dt}{-7} \\\\\int\limits {\frac{-(4-7x)+4}{\sqrt{4-7x} } } \, dx=\int\limits {\frac{-(t)+4}{-7\sqrt{t} } } \, dt = \int\limits {\frac{t-4}{7\sqrt{t} } } \, dt=\frac{1}{7}(\int\limits {\frac{t-4}{\sqrt{t} } } \, dt) = \frac{1}{7}(\int\limits{(\frac{t}{\sqrt{t} }-\frac{4}{\sqrt{t}})}\,dt)=$

$=\frac{1}{7}(\int\limits{(\sqrt{t} })}\,dt-4\int\limits{(\frac{1}{\sqrt{t} }})}\,dt) = \frac{1}{7}(\frac{2\sqrt{t^3} }{3} -8\sqrt{t} +C) = \frac{2t\sqrt{t} }{21} -\frac{8\sqrt{t}}{7} +C = \\\\=\frac{2(4-7x)\sqrt{4-7x} }{21} -\frac{8\sqrt{4-7x}}{7} +C$

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