y^'= x/(1+x^2)

Lösung

y^{^{'}} = \frac{x}{1 + x ^{2}}

y = \frac{1}{2} ln (1 + x^{2}) + c_{1}

Schritte zur Lösung

y^{'} (x) = \frac{x}{1 + x ^{2}}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int \frac{x}{1 + x ^{2}} d x

\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (1 + x^{2}) + c_{1}

y = \frac{1}{2} ln (1 + x^{2}) + c_{1}

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{x})

y^{^{'}} = y^{2} - (x y)^{2}

\int \frac{1}{3 x ( 2 + x )} d x

x \to 2 - lim (arcsin (\frac{x}{2}))

x \to - 1 lim ((x + 2 x^{4})^{3})