What is the integral from 0 to 1 of 1/(t+1) ?

The integral from 0 to 1 of 1/(t+1) is ln(2)

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integral from 0 to 1 of 1/(t+1)

\int_{0}^{1} \frac{1}{t + 1} d t

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{1} \frac{1}{t + 1} d t

Apply u-substitution

: \int_{1}^{2} \frac{1}{u} d u

= \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\frac{d y}{d x}, x y = 36

\frac{d y}{d t} + 2 t y = 2 t

\int \frac{1}{x ^{2} ( e ^{\frac{1}{x}} - 5 )} d x

\int \frac{1}{2} x^{2} - 2 x + 6 d x

\frac{3}{s} + \frac{24}{s ^{2}}

What is the integral from 0 to 1 of 1/(t+1) ?
The integral from 0 to 1 of 1/(t+1) is ln(2)