What is the integral from 0 to 1 of (1+2^x)^22^{-x} ?

The integral from 0 to 1 of (1+2^x)^22^{-x} is 3/(2ln(2))+2

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integral from 0 to 1 of (1+2^x)^22^{-x}

\int_{0}^{1} (1 + 2^{x})^{2} 2^{- x} d x

\frac{3}{2 ln ( 2 )} + 2

Decimal Notation

4.16404 \dots

Solution steps

\int_{0}^{1} (1 + 2^{x})^{2} \cdot 2^{- x} d x

Expand

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

= \int_{0}^{1} 2^{- x} + 2 + 2^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 2^{- x} d x + \int_{0}^{1} 2 d x + \int_{0}^{1} 2^{x} d x

\int_{0}^{1} 2^{- x} d x = \frac{1}{2 ln ( 2 )}

\int_{0}^{1} 2 d x = 2

\int_{0}^{1} 2^{x} d x = \frac{1}{ln ( 2 )}

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

Simplify

\frac{1}{2 ln ( 2 )} + 2 + \frac{1}{ln ( 2 )} : \frac{3}{2 ln ( 2 )} + 2

= \frac{3}{2 ln ( 2 )} + 2

\frac{\partial}{\partial x} (\cdot e)

f (x) = (x^{2} + 1) (2 - x)

x \to - 3 lim (2 x + 5)

\frac{d}{d x} (cos (x^{- 6}))

What is the integral from 0 to 1 of (1+2^x)^22^{-x} ?
The integral from 0 to 1 of (1+2^x)^22^{-x} is 3/(2ln(2))+2