integral of (1/3 e^{x+3}+3/x-5)

Lösung

\int (\frac{1}{3} e^{x + 3} + \frac{3}{x} - 5) d x

\frac{1}{3} e^{3 + x} + 3 ln ∣ x ∣ - 5 x + C

Schritte zur Lösung

\int \frac{1}{3} e^{x + 3} + \frac{3}{x} - 5 d x

Wende die Summenregel an:

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

= \int \frac{1}{3} e^{x + 3} d x + \int \frac{3}{x} d x - \int 5 d x

\int \frac{1}{3} e^{x + 3} d x = \frac{1}{3} e^{3 + x}

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int 5 d x = 5 x

= \frac{1}{3} e^{3 + x} + 3 ln ∣ x ∣ - 5 x

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} e^{3 + x} + 3 ln ∣ x ∣ - 5 x + C

\int \frac{x ^{2} - 11 x - 18}{x ( x ^{2} + 3 x + 3 )} d x

\frac{\partial}{\partial x} (5 x y - 4 x^{2} y)

\frac{\partial}{\partial x} (x)

x \to 0 lim (sin^{4} (x))

l a pl a ce t r an s f or m cos (2 (t - \frac{5 π}{2}))