What is the integral from 0 to 3 of (e^x)/(4+e^x) ?

The integral from 0 to 3 of (e^x)/(4+e^x) is ln(4+e^3)-ln(5)

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integral from 0 to 3 of (e^x)/(4+e^x)

\int_{0}^{3} \frac{e ^{x}}{4 + e ^{x}} d x

ln (4 + e^{3}) - ln (5)

Decimal

1.57217 \dots

Solution steps

\int_{0}^{3} \frac{e ^{x}}{4 + e ^{x}} d x

Apply u-substitution

: \int_{5}^{4 + e^{3}} \frac{1}{u} d u

= \int_{5}^{4 + e^{3}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{5}^{4 + e^{3}}

Compute the boundaries:

ln (4 + e^{3}) - ln (5)

= ln (4 + e^{3}) - ln (5)

\int_{0}^{3} 3 y - y^{2} d y

\int_{0}^{π} x^{2} cos (\frac{x}{2}) d x

\int_{- 2}^{2} (∣ x ∣) - (x^{2} - 2) d x

\int_{0}^{2 x} cos (x^{2}) d x

\int_{0}^{t} u e^{2 u} cos (3 u) d u

What is the integral from 0 to 3 of (e^x)/(4+e^x) ?
The integral from 0 to 3 of (e^x)/(4+e^x) is ln(4+e^3)-ln(5)