integral of sqrt(1-e^{2x)}

Lösung

\int 1 - e^{2 x} d x

- \frac{1}{2} ln 1 - e^{2 x} + 1 + \frac{1}{2} ln 1 - e^{2 x} - 1 + 1 - e^{2 x} + C

Schritte zur Lösung

\int 1 - e^{2 x} d x

Wende U-Substitution an

= \int - \frac{u}{2 ( - u + 1 )} d u

Entferne die Konstante:

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

= - \frac{1}{2} \cdot \int \frac{u}{- u + 1} d u

Wende U-Substitution an

= - \frac{1}{2} \cdot \int \frac{2 v ^{2}}{- v ^{2} + 1} d v

Entferne die Konstante:

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

= - \frac{1}{2} \cdot 2 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= - \frac{1}{2} \cdot 2 \cdot \int \frac{1}{- v ^{2} + 1} - 1 d v

Wende die Summenregel an:

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

= - \frac{1}{2} \cdot 2 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v)

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

\int 1 d v = v

= - \frac{1}{2} \cdot 2 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

Ersetze zurück

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

Vereinfache

- \frac{1}{2} \cdot 2 (\frac{ln 1 - e ^{2 x} + 1}{2} - \frac{ln 1 - e ^{2 x} - 1}{2} - 1 - e^{2 x}) : - \frac{1}{2} ln 1 - e^{2 x} + 1 + \frac{1}{2} ln 1 - e^{2 x} - 1 + 1 - e^{2 x}

= - \frac{1}{2} ln 1 - e^{2 x} + 1 + \frac{1}{2} ln 1 - e^{2 x} - 1 + 1 - e^{2 x}

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{2} ln 1 - e^{2 x} + 1 + \frac{1}{2} ln 1 - e^{2 x} - 1 + 1 - e^{2 x} + C

x \to 0 lim (\frac{3}{x} - \frac{2}{ln ( x )})

\int 2 π (4 x^{2} + 4 x + 1) d x

x \to \frac{2}{3} lim (\frac{27 x ^{3} - 8}{9 x ^{2} - 4})

d er i v a t i v e f (x) = \frac{cos ( x )}{x ^{3}}

x \to 5 lim (2 x^{2} - 7 x + 13)