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

Lösung

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

\frac{- x ^{2} - 6 x - 12}{2 e ^{\frac{1}{2} (x + 1)}} + C

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= \frac{1}{4} \cdot \int \frac{u ^{2} - 1}{e ^{\frac{u}{2}}} d u

Wende U-Substitution an

= \frac{1}{4} \cdot \int \frac{2 ( 4 v ^{2} - 1 )}{e ^{v}} d v

Entferne die Konstante:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{4 v ^{2} - 1}{e ^{v}} d v

Multipliziere aus

\frac{4 v ^{2} - 1}{e ^{v}} : \frac{4 v ^{2}}{e ^{v}} - \frac{1}{e ^{v}}

= \frac{1}{4} \cdot 2 \cdot \int \frac{4 v ^{2}}{e ^{v}} - \frac{1}{e ^{v}} 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}{4} \cdot 2 (\int \frac{4 v ^{2}}{e ^{v}} d v - \int \frac{1}{e ^{v}} d v)

\int \frac{4 v ^{2}}{e ^{v}} d v = 4 (- \frac{v ^{2}}{e ^{v}} + 2 (- \frac{v}{e ^{v}} - \frac{1}{e ^{v}}))

\int \frac{1}{e ^{v}} d v = - \frac{1}{e ^{v}}

= \frac{1}{4} \cdot 2 (4 (- \frac{v ^{2}}{e ^{v}} + 2 (- \frac{v}{e ^{v}} - \frac{1}{e ^{v}})) - (- \frac{1}{e ^{v}}))

Ersetze zurück

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

Vereinfache

\frac{1}{4} \cdot 2 (4 (- \frac{( \frac{x + 1}{2} ) ^{2}}{e ^{\frac{x + 1}{2}}} + 2 (- \frac{\frac{x + 1}{2}}{e ^{\frac{x + 1}{2}}} - \frac{1}{e ^{\frac{x + 1}{2}}})) - (- \frac{1}{e ^{\frac{x + 1}{2}}})) : \frac{- x ^{2} - 6 x - 12}{2 e ^{\frac{1}{2} (x + 1)}}

= \frac{- x ^{2} - 6 x - 12}{2 e ^{\frac{1}{2} (x + 1)}}

Füge eine Konstante zur Lösung hinzu

= \frac{- x ^{2} - 6 x - 12}{2 e ^{\frac{1}{2} (x + 1)}} + C

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