f^'(x)=-2ln(x)+x^2-4+e^x+x^e

Lösung

f^{^{'}} (x) = - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e}

f (x) = \frac{x ^{3}}{3} - 2 x ln (x) - 2 x + \frac{1}{e + 1} x^{e + 1} + e^{x} + c_{1}

Schritte zur Lösung

f^{'} (x) = - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e}

Wenn

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

dann

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

f (x) = \int - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e} d x

\int - 2 ln (x) + x^{2} - 4 + e^{x} + x^{e} d x = \frac{x ^{3}}{3} - 2 x ln (x) - 2 x + \frac{1}{e + 1} x^{e + 1} + e^{x} + c_{1}

f (x) = \frac{x ^{3}}{3} - 2 x ln (x) - 2 x + \frac{1}{e + 1} x^{e + 1} + e^{x} + c_{1}

a re a f (x) = x^{2} - 6 x + 8, g (x) = - 2 x + 8

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

x \to 0 lim (\frac{x}{x - 1})

x \to 0 lim (x - 4)

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