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integral of sqrt(2+e^x)

Lösung

\int 2 + e^{x} d x

- 2 ln \frac{1}{2} (2 + e^{x}) + 1 + 2 ln \frac{1}{2} (2 + e^{x}) - 1 + 22 \frac{1}{2} (2 + e^{x}) + C

Schritte zur Lösung

\int 2 + e^{x} d x

Wende U-Substitution an

= \int \frac{2 + u}{u} d u

Wende U-Substitution an

= \int \frac{2 v ^{2}}{v ^{2} - 2} d v

Entferne die Konstante:

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

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

Wende integrale Substitution an

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

Entferne die Konstante:

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

= 22 \cdot \int \frac{w ^{2}}{w ^{2} - 1} d w

Faktorisiere

w^{2} - 1 : - (- w^{2} + 1)

= 22 \cdot \int \frac{w ^{2}}{- ( - w ^{2} + 1 )} d w

Entferne die Konstante:

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

= 22 (- \int \frac{w ^{2}}{- w ^{2} + 1} d w)

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

= 22 (- \int \frac{1}{- w ^{2} + 1} - 1 d w)

Wende die Summenregel an:

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

= 22 (- (\int \frac{1}{- w ^{2} + 1} d w - \int 1 d w))

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

\int 1 d w = w

= 22 (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2} - w))

Ersetze zurück

= 22 - \frac{ln \frac{2 + e ^{x}}{2} + 1}{2} - \frac{ln \frac{2 + e ^{x}}{2} - 1}{2} - \frac{2 + e ^{x}}{2}

Vereinfache

22 - \frac{ln \frac{2 + e ^{x}}{2} + 1}{2} - \frac{ln \frac{2 + e ^{x}}{2} - 1}{2} - \frac{2 + e ^{x}}{2} : - 2 ln \frac{1}{2} (2 + e^{x}) + 1 + 2 ln \frac{1}{2} (2 + e^{x}) - 1 + 22 \frac{1}{2} (2 + e^{x})

= - 2 ln \frac{1}{2} (2 + e^{x}) + 1 + 2 ln \frac{1}{2} (2 + e^{x}) - 1 + 22 \frac{1}{2} (2 + e^{x})

Füge eine Konstante zur Lösung hinzu

= - 2 ln \frac{1}{2} (2 + e^{x}) + 1 + 2 ln \frac{1}{2} (2 + e^{x}) - 1 + 22 \frac{1}{2} (2 + e^{x}) + C