What is the integral from 0 to 1 of (5^x)/(5^x+1) ?

The integral from 0 to 1 of (5^x)/(5^x+1) is (ln(3))/(ln(5))

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integral from 0 to 1 of (5^x)/(5^x+1)

\int_{0}^{1} \frac{5 ^{x}}{5 ^{x} + 1} d x

\frac{ln ( 3 )}{ln ( 5 )}

Decimal Notation

0.68260 \dots

Solution steps

\int_{0}^{1} \frac{5 ^{x}}{5 ^{x} + 1} d x

Apply u-substitution

: \int_{2}^{6} \frac{1}{ln ( 5 ) u} d u

= \int_{2}^{6} \frac{1}{ln ( 5 ) u} d u

Take the constant out:

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

= \frac{1}{ln ( 5 )} \cdot \int_{2}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 5 )} [ln ∣ u ∣]_{2}^{6}

Compute the boundaries:

ln (6) - ln (2)

= \frac{1}{ln ( 5 )} (ln (6) - ln (2))

Simplify

= \frac{ln ( 3 )}{ln ( 5 )}

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What is the integral from 0 to 1 of (5^x)/(5^x+1) ?
The integral from 0 to 1 of (5^x)/(5^x+1) is (ln(3))/(ln(5))