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integral of (sqrt(4+2x))/x

Lösung

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

- 4 (\frac{1}{2} ln \frac{4 + 2 x}{2} + 1 - \frac{1}{2} ln \frac{4 + 2 x}{2} - 1 - \frac{1}{2} 4 + 2 x) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{u}{u - 4} d u

Wende U-Substitution an

= \int \frac{2 v ^{2}}{v ^{2} - 4} 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} - 4} d v

Faktorisiere

v^{2} - 4 : - (- v^{2} + 4)

= 2 \cdot \int \frac{v ^{2}}{- ( - v ^{2} + 4 )} d v

Entferne die Konstante:

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

= 2 (- \int \frac{v ^{2}}{- v ^{2} + 4} d v)

Wende integrale Substitution an

= 2 (- \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

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

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

= 2 (- 2 \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

= 2 (- 2 (\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

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

Ersetze zurück

= 2 - 2 \frac{ln \frac{4 + 2 x}{2} + 1}{2} - \frac{ln \frac{4 + 2 x}{2} - 1}{2} - \frac{4 + 2 x}{2}

Vereinfache

2 - 2 \frac{ln \frac{4 + 2 x}{2} + 1}{2} - \frac{ln \frac{4 + 2 x}{2} - 1}{2} - \frac{4 + 2 x}{2} : - 4 (\frac{1}{2} ln \frac{4 + 2 x}{2} + 1 - \frac{1}{2} ln \frac{4 + 2 x}{2} - 1 - \frac{1}{2} 4 + 2 x)

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

Füge eine Konstante zur Lösung hinzu

= - 4 (\frac{1}{2} ln \frac{4 + 2 x}{2} + 1 - \frac{1}{2} ln \frac{4 + 2 x}{2} - 1 - \frac{1}{2} 4 + 2 x) + C

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