What is the integral from 0 to 3 of (9e^{9x})/(1+e^{9x)} ?

The integral from 0 to 3 of (9e^{9x})/(1+e^{9x)} is ln(1+e^{27})-ln(2)

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integral from 0 to 3 of (9e^{9x})/(1+e^{9x)}

\int_{0}^{3} \frac{9 e ^{9 x}}{1 + e ^{9 x}} d x

ln (1 + e^{27}) - ln (2)

Decimal

26.30685 \dots

Solution steps

\int_{0}^{3} \frac{9 e ^{9 x}}{1 + e ^{9 x}} d x

Take the constant out:

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

= 9 \cdot \int_{0}^{3} \frac{e ^{9 x}}{1 + e ^{9 x}} d x

Apply u-substitution

: \int_{2}^{1 + e^{27}} \frac{1}{9 u} d u

= 9 \cdot \int_{2}^{1 + e^{27}} \frac{1}{9 u} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int_{2}^{1 + e^{27}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 9 \cdot \frac{1}{9} [ln ∣ u ∣]_{2}^{1 + e^{27}}

Simplify

9 \cdot \frac{1}{9} [ln ∣ u ∣]_{2}^{1 + e^{27}} : [ln ∣ u ∣]_{2}^{1 + e^{27}}

= [ln ∣ u ∣]_{2}^{1 + e^{27}}

Compute the boundaries:

ln (1 + e^{27}) - ln (2)

= ln (1 + e^{27}) - ln (2)

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What is the integral from 0 to 3 of (9e^{9x})/(1+e^{9x)} ?
The integral from 0 to 3 of (9e^{9x})/(1+e^{9x)} is ln(1+e^{27})-ln(2)