What is the integral from 0 to 1 of (2^x)/(2^x+2) ?

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

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integral from 0 to 1 of (2^x)/(2^x+2)

\int_{0}^{1} \frac{2 ^{x}}{2 ^{x} + 2} d x

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

Decimal Notation

0.41503 \dots

Solution steps

\int_{0}^{1} \frac{2 ^{x}}{2 ^{x} + 2} d x

Apply u-substitution

: \int_{3}^{4} \frac{1}{ln ( 2 ) u} d u

= \int_{3}^{4} \frac{1}{ln ( 2 ) u} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int_{3}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 2 )} [ln ∣ u ∣]_{3}^{4}

Compute the boundaries:

2 ln (2) - ln (3)

= \frac{1}{ln ( 2 )} (2 ln (2) - ln (3))

Simplify

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

\int tan^{3} (\frac{x}{2}) sec^{2} (\frac{x}{2}) d x

\frac{d}{d x} (tan (x + y))

\frac{d}{d x} (\frac{( 5 - x ) ^{- 2}}{2})

\frac{d}{d x} (0.2 x^{2})

\frac{d ^{2}}{d x ^{2}} (14 x^{3} - 36 x^{2})

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