Question

Можете мне решить этот неопределенный интеграл?
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Answer 1

$\int \frac{\sqrt{1+\sqrt{x}}}{x\sqrt[4]{x^3} } dx = \int ((1+x^{\frac{1}{2}})^\frac{1}{2}*x^{-\frac{7}{4}})dx = \int (x^{-\frac{7}{4}}*(x^{\frac{1}{2}}+1)^\frac{1}{2})dx; \\|m=-\frac{7}{4}\\| a = 1\\|n = \frac{1}{2}\\ |b = 1\\|p=\frac{1}{2}\\\frac{m+1}{n} + p = -1 \in Z = t^2 = 1 + x^{-\frac{1}{2}} = x^{-\frac{1}{2}} = t^2-1 = x^{-1} = (t^2-1)^2 = x = (t^2-1)^{-2} = dx = -2(t^2-1)^{-3}*2tdt = -4t(t^2-1)^{-3}dt\\$

$\int \frac{\sqrt{1+\sqrt{x}}}{x\sqrt[4]{x^3} } dx = \int \frac{\sqrt{1+\sqrt{(t^2-1)^{-2}}}}{(t^2-1)^{-2*\frac{7}{4} }} * -4t(t^2-1)^{-3}dt = \int \frac{\sqrt{1 + \frac{1}{t^2-1}}}{(t^2-1)^{-\frac{7}{2} }} *-4t(t^2-1)^{-3}dt =$

$= \int -\frac{\sqrt{\frac{t^2-1+1}{t^2-1}}}{(t^2-1)^{-\frac{7}{2}}} *4t(t^2-1)^{-3}dt = \int- \frac{\sqrt{\frac{t^2}{t^2-1}}}{(t^2-1)^{-3}*(t^2-1)^{-\frac{1}{2}}} * 4t (t^2-1)^{-3}dt = \int- 4t^2dt = -\frac{4}{3}t^3 + c = - \frac{4}{3}(1+x^{-\frac{1}{2}})^{\frac{3}{2}} + c = -\frac{4}{3}(1+\frac{1}{\sqrt{x}} )^{\frac{3}{2}} + c$

Answer 2

$-\frac{4}{3}\sqrt{\left(1+x^{-\frac{1}{2}}\right)^3}+C$

Пошаговое объяснение: