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

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

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

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

ln (e + 1)

Decimal Notation

1.31326 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e + 1}

Compute the boundaries:

ln (e + 1)

= ln (e + 1)

y^{^{'}} = y sin (x), y (0) = 1

\frac{1}{6} y^{^{''}} + 6 y = tan (6 t) - \frac{1}{2} e^{3 t}

\frac{\partial}{\partial x} (tan (y))

\frac{d y}{d x} = \frac{( 2880 - y )}{320}

y^{^{'}} + y = 1, y (0) = 1

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