What is the integral of (20e^{0.02x})/(1+e^{0.02x)} ?

The integral of (20e^{0.02x})/(1+e^{0.02x)} is 1000ln|1+e^{0.02x}|+C

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integral of (20e^{0.02x})/(1+e^{0.02x)}

\int \frac{20 e ^{0.02 x}}{1 + e ^{0.02 x}} d x

1000 ln 1 + e^{0.02 x} + C

Solution steps

\int \frac{20 e ^{0.02 x}}{1 + e ^{0.02 x}} d x

Take the constant out:

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

= 20 \cdot \int \frac{e ^{0.02 x}}{1 + e ^{0.02 x}} d x

Apply u-substitution

: \int \frac{1}{0.02 u} d u

= 20 \cdot \int \frac{1}{0.02 u} d u

Take the constant out:

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

= 20 \cdot \frac{1}{0.02} \cdot \int \frac{1}{u} d u

Simplify

= 20 \cdot 50 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 20 \cdot 50 ln ∣ u ∣

Substitute back

u = 1 + e^{0.02 x}

= 20 \cdot 50 ln 1 + e^{0.02 x}

Simplify

= 1000 ln 1 + e^{0.02 x}

Add a constant to the solution

= 1000 ln 1 + e^{0.02 x} + C

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What is the integral of (20e^{0.02x})/(1+e^{0.02x)} ?
The integral of (20e^{0.02x})/(1+e^{0.02x)} is 1000ln|1+e^{0.02x}|+C