(dy)/(dx)=x^3+2x+pi

Lösung

\frac{d y}{d x} = x^{3} + 2 x + π

y = \frac{x ^{4}}{4} + x^{2} + π x + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = x^{3} + 2 x + π

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = x^{3} + 2 x + π

Wenn

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

dann

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

y = \int x^{3} + 2 x + π d x

\int x^{3} + 2 x + π d x = \frac{x ^{4}}{4} + x^{2} + π x + c_{1}

y = \frac{x ^{4}}{4} + x^{2} + π x + c_{1}

\int_{- 1}^{2} (- x^{2} + x + 2) d x

\frac{d}{d x} (tan (x^{3}))

d er i v a t i v e \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

x \to 4 + lim (2 x - 7)

s l o p e x + \frac{1}{x ^{3}}