What is the integral from 1 to 3 of (2^x)/(2^x+4) ?

The integral from 1 to 3 of (2^x)/(2^x+4) is 1

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integral from 1 to 3 of (2^x)/(2^x+4)

\int_{1}^{3} \frac{2 ^{x}}{2 ^{x} + 4} d x

1

Solution steps

\int_{1}^{3} \frac{2 ^{x}}{2 ^{x} + 4} d x

Apply u-substitution

= \int_{6}^{12} \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_{6}^{12} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 2 )} [ln ∣ u ∣]_{6}^{12}

Compute the boundaries:

ln (12) - ln (6)

= \frac{1}{ln ( 2 )} (ln (12) - ln (6))

Simplify

= 1

What is the integral from 1 to 3 of (2^x)/(2^x+4) ?
The integral from 1 to 3 of (2^x)/(2^x+4) is 1