What is the integral from 0 to 4 of (e^x)/(10+e^x) ?

The integral from 0 to 4 of (e^x)/(10+e^x) is ln(10+e^4)-ln(11)

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integral from 0 to 4 of (e^x)/(10+e^x)

\int_{0}^{4} \frac{e ^{x}}{10 + e ^{x}} d x

ln (10 + e^{4}) - ln (11)

Decimal Notation

1.77029 \dots

Solution steps

\int_{0}^{4} \frac{e ^{x}}{10 + e ^{x}} d x

Apply u-substitution

: \int_{11}^{10 + e^{4}} \frac{1}{u} d u

= \int_{11}^{10 + e^{4}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{11}^{10 + e^{4}}

Compute the boundaries:

ln (10 + e^{4}) - ln (11)

= ln (10 + e^{4}) - ln (11)

\int_{- 1}^{0} 4 d x

\int_{\frac{π}{6}}^{\frac{π}{4}} sin (t) d t

\int_{0}^{\frac{1}{3}} \frac{1}{1 + 9 x ^{2}} d x

\int_{- 1}^{1} \frac{e ^{x} - e ^{- x}}{2} d x

\int_{0}^{3} 1 - x^{2} d x

What is the integral from 0 to 4 of (e^x)/(10+e^x) ?
The integral from 0 to 4 of (e^x)/(10+e^x) is ln(10+e^4)-ln(11)