integral of 4x^{-1}-4^x+10e^{2x}-8x

Lösung

\int 4 x^{- 1} - 4^{x} + 10 e^{2 x} - 8 x d x

4 ln ∣ x ∣ - \frac{2 ^{2 x - 1}}{ln ( 2 )} + 5 e^{2 x} - 4 x^{2} + C

Schritte zur Lösung

\int 4 x^{- 1} - 4^{x} + 10 e^{2 x} - 8 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 4 x^{- 1} d x - \int 4^{x} d x + \int 10 e^{2 x} d x - \int 8 x d x

\int 4 x^{- 1} d x = 4 ln ∣ x ∣

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

\int 10 e^{2 x} d x = 5 e^{2 x}

\int 8 x d x = 4 x^{2}

= 4 ln ∣ x ∣ - \frac{2 ^{2 x - 1}}{ln ( 2 )} + 5 e^{2 x} - 4 x^{2}

Füge eine Konstante zur Lösung hinzu

= 4 ln ∣ x ∣ - \frac{2 ^{2 x - 1}}{ln ( 2 )} + 5 e^{2 x} - 4 x^{2} + C

\frac{d}{d x} ((x - 7)^{2})

\int x^{3} \cdot e^{- x^{2}} d x

x \to \infty lim (arctan (ln (x)))

\frac{d ^{3}}{d x ^{3}} (\frac{6}{x})

\int \frac{e ^{x}}{1 + 2 e ^{x}} d x