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integral of sqrt(10-e^{2x)}

Lösung

\int 10 - e^{2 x} d x

- 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x})) + C

Schritte zur Lösung

\int 10 - e^{2 x} d x

Wende U-Substitution an

= \int \frac{10 - u}{2 u} 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{10 - u}{u} d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int - \frac{2 v ^{2}}{- v ^{2} + 10} d v

Entferne die Konstante:

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

= \frac{1}{2} (- 2 \cdot \int \frac{v ^{2}}{- v ^{2} + 10} d v)

Wende integrale Substitution an

= \frac{1}{2} (- 2 \cdot \int \frac{10 w ^{2}}{- w ^{2} + 1} d w)

Entferne die Konstante:

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

= \frac{1}{2} (- 210 \cdot \int \frac{w ^{2}}{- w ^{2} + 1} d w)

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

= \frac{1}{2} (- 210 \cdot \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

= \frac{1}{2} (- 210 (\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

= \frac{1}{2} (- 210 (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2} - w))

Ersetze zurück

= \frac{1}{2} - 210 \frac{ln \frac{10 - e ^{2 x}}{10} + 1}{2} - \frac{ln \frac{10 - e ^{2 x}}{10} - 1}{2} - \frac{10 - e ^{2 x}}{10}

Vereinfache

\frac{1}{2} - 210 \frac{ln \frac{10 - e ^{2 x}}{10} + 1}{2} - \frac{ln \frac{10 - e ^{2 x}}{10} - 1}{2} - \frac{10 - e ^{2 x}}{10} : - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x}))

= - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x}))

Füge eine Konstante zur Lösung hinzu

= - 10 (\frac{1}{2} ln \frac{10 - e ^{2 x}}{10} + 1 - \frac{1}{2} ln \frac{10 - e ^{2 x}}{10} - 1 - \frac{1}{10} (10 - e^{2 x})) + C

\frac{d}{d x} (2 + 2 + x)

\frac{d P}{d t} = Pt, P (1) = 3

\int - x^{2} + 9 x d x

y^{^{''}} - 2 y^{^{'}} - 36 y = 0

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