Question

Найти интеграл 60 баллов

Answer 1

Ответ:

Пошаговое объяснение:

$\int {\sqrt[3]{\frac{1-x}{1+x} } } \, \frac{dx}{x} =\begin{Vmatrix} \frac{1-x}{1+x}=t^3, \; 1-x=t^3(1+x), \; 1-x=t^3+xt^3\\x+xt^3=1-t^3, \; x(1+t^3)=1-t^3, \; x=\frac{1-t^3}{1+t^3}\\dx=\frac{(1-t^3)'*(1+t^3)-(1-t^3)*(1+t^3)'}{(1+t^3)^2}dt=\\=\frac{-3t^2(1+t^3)-3t^2(1-t^3)}{(1+t^3)^2}dt=\frac{-3t^2(1+t^3+1-t^3)}{(1+t^3)^2}dt\\dx=\frac{-6t^2}{(1+t^3)^2} dt\end{Vmatrix}=\int {t*\frac{1+t^3}{1-t^3}*\frac{-6t^2}{(1+t^3)^2} } \,dt=$$=-6\int {\frac{t^3}{(1-t^3)(1+t^3)} } \, dt=-6\int {\frac{t^3}{(1-t)(1+t+t^2)(1+t)(1-t+t^2)} } \, dt=\\-6\int {\frac{1}{6}(\frac{1}{1-t}-\frac{1}{1+t}+\frac{t+2}{1+t+t^2}+\frac{t-2}{1-t+t^2}) } \, dt =\\-\int {\frac{dt}{1-t} } \, +\int {\frac{dt}{1+t} } \,-\int {\frac{t+2}{t^2+t+1} } \, dt -\int {\frac{t-2}{t^2-t+1} } \, dt=\\-\int {\frac{dt}{1-t} } \, +\int {\frac{dt}{1+t} } \,-\frac{1}{2}\int {\frac{(2t+1)+3}{t^2+t+1} } \, dt-\frac{1}{2}\int {\frac{(2t-1)-3}{t^2-t+1} } \, dt=$

$-\int {\frac{dt}{1-t} } \, +\int {\frac{dt}{1+t} } \,-\frac{1}{2} \int {\frac{2t+1}{t^2+t+1} } \, -\frac{1}{2}\int {\frac{3}{t^2+2*t*0.5+0.5^2+\frac{3}{4} } } \,dt-\\\frac{1}{2}\int {\frac{2t-1}{t^2-t+1} } \, dt+\frac{1}{2}\int {\frac{3}{t^2-2*t*0.5+0.5^2+\frac{3}{4} } } \, dt=$$\int {\frac{d(1-t)}{1-t} } \, +\int {\frac{d(1+t)}{1+t} } \,-\frac{1}{2} \int {\frac{d(t^2+t+1)}{t^2+t+1} } \, -\\\frac{3}{2}\int {\frac{d(t+0.5)}{(t+0.5)^2+(\frac{\sqrt{3} }{2} )^2} } \,-\frac{1}{2}\int {\frac{d(t^2-t+1)}{t^2-t+1} } \, +\frac{3}{2}\int {\frac{d(t-0.5)}{(t-0.5)^2+(\frac{\sqrt{3} }{2})^2 } } \,=$$\ln |1-t|+\ln |1+t|-\frac{1}{2}\ln |t^2+t+1|-\frac{3}{2}*\frac{2}{\sqrt{3} } \arctan2\frac{t+0.5}{\sqrt{3} }-\frac{1}{2}\ln |t^2-t+1|+\frac{3}{2}*\frac{2}{\sqrt{3} } \arctan2\frac{t-0.5}{\sqrt{3} }+C=\\\ln |1-t^2|-\frac{1}{2}\ln |(t^2+t+1)(t^2-t+1)|-\sqrt{3} (\arctan\frac{2t+1}{\sqrt{3} }- \arctan\frac{2t-1}{\sqrt{3} })+C=$

$\ln |1-(\sqrt[3]{\frac{1-x}{1+x} })^2 |-\frac{1}{2}\ln |(\sqrt[3]{\frac{1-x}{1+x} })^2+\sqrt[3]{\frac{1-x}{1+x} }+1)(\sqrt[3]{\frac{1-x}{1+x} })^2-\sqrt[3]{\frac{1-x}{1+x} }+1)|-\\\sqrt{3}(\arctan \frac{2\sqrt[3]{\frac{1-x}{1+x} }+1 }{\sqrt{3} } -\arctan \frac{2\sqrt[3]{\frac{1-x}{1+x} }-1 }{\sqrt{3} } )+C$