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Beliebt Rechnen >

integral of 1/(4^x)

Lösung

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

Lösung

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

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{1}{ln ( 2 ) \cdot 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} d u

Entferne die Konstante:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} d u

\frac{1}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} = \frac{1 - 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} + 1

= \frac{1}{ln ( 2 )} \cdot \int \frac{1 - 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} + 1 d u

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}{ln ( 2 )} (\int \frac{1 - 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} d u + \int 1 d u)

\int \frac{1 - 2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}} u} d u = \frac{- 1 - 2 ^{\frac{2 l n ( u ) + l n ( 2 )}{l n ( 2 )}}}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}}

\int 1 d u = u

= \frac{1}{ln ( 2 )} (\frac{- 1 - 2 ^{\frac{2 l n ( u ) + l n ( 2 )}{l n ( 2 )}}}{2 ^{\frac{l n ( u ) + l n ( 2 )}{l n ( 2 )}}} + u)

Setze in

u = 4^{x}

ein

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

Vereinfache

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

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

Füge eine Konstante zur Lösung hinzu

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

Graph

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