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

Lösung

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

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

Schritte zur Lösung

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

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

\int ln (3) d x = ln (3) x

\int \frac{1}{x} d x = ln ∣ x ∣

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

Füge eine Konstante zur Lösung hinzu

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

\frac{d y}{d x} = - \frac{( 2 x ^{2} + y )}{( x ^{2} y - x )}

\frac{\partial}{\partial x} (e^{t} cos (x))

d er i v a t i v e 7 x^{3} + 5 x^{2} + 4 x

\frac{\partial}{\partial x} (cos (x + 6 y))

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