integral of (e^2x)/(4+e^4x)

Lösung

\int \frac{e ^{2} x}{4 + e ^{4} x} d x

\frac{1}{e ^{6}} (4 + e^{4} x - 4 ln 4 + e^{4} x) + C

Schritte zur Lösung

\int \frac{e ^{2} x}{4 + e ^{4} x} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{2} \cdot \int \frac{x}{4 + e ^{4} x} d x

Wende U-Substitution an

= e^{2} \cdot \int \frac{u - 4}{e ^{8} u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e^{2} \frac{1}{e ^{8}} \cdot \int \frac{u - 4}{u} d u

Multipliziere aus

\frac{u - 4}{u} : 1 - \frac{4}{u}

= e^{2} \frac{1}{e ^{8}} \cdot \int 1 - \frac{4}{u} d u

Wende die Summenregel an:

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

= e^{2} \frac{1}{e ^{8}} (\int 1 d u - \int \frac{4}{u} d u)

\int 1 d u = u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= e^{2} \frac{1}{e ^{8}} (u - 4 ln ∣ u ∣)

Setze in

u = 4 + e^{4} x

ein

= e^{2} \frac{1}{e ^{8}} (4 + e^{4} x - 4 ln 4 + e^{4} x)

Vereinfache

e^{2} \frac{1}{e ^{8}} (4 + e^{4} x - 4 ln 4 + e^{4} x) : \frac{1}{e ^{6}} (4 + e^{4} x - 4 ln 4 + e^{4} x)

= \frac{1}{e ^{6}} (4 + e^{4} x - 4 ln 4 + e^{4} x)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{e ^{6}} (4 + e^{4} x - 4 ln 4 + e^{4} x) + C

\frac{d y}{d x} = \frac{( 2 x + 1 )}{2 y}

d er i v a t i v e 9 x (2 x^{2} + 1)

s l o p e 2 x

\int 9 t^{2} + 4 d t

\frac{\partial}{\partial x} (\frac{9}{x ^{2} + y ^{2}})