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integral of (sqrt(1+4x^2))/x

Lösung

\int \frac{1 + 4 x ^{2}}{x} d x

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

Schritte zur Lösung

\int \frac{1 + 4 x ^{2}}{x} d x

Wende U-Substitution an

= \int \frac{1 + 4 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{1 + 4 u}{u} 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

Faktorisiere

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

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{- 2}{7 x ^{4}} d x

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\int (x^{5} - x^{3} + x) e^{- x} d x