What is the integral of 2^{x+1} ?

The integral of 2^{x+1} is (2^{x+1})/(ln(2))+C

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integral of 2^{x+1}

\int 2^{x + 1} d x

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

Solution steps

\int 2^{x + 1} d x

2^{x + 1} = 2^{1} \cdot 2^{x}

= \int 2^{1} \cdot 2^{x} d x

Take the constant out:

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

= 2^{1} \cdot \int 2^{x} d x

Simplify

= 2 \cdot \int 2^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

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

Simplify

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

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

Add a constant to the solution

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

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

\int sin^{4} (u) d u

y = \frac{a x + b}{c x + d}

\int n e^{- n} d n

f (x) = - 5 ln (sin (x))

What is the integral of 2^{x+1} ?
The integral of 2^{x+1} is (2^{x+1})/(ln(2))+C