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

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

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integral from 0 to 1 of (-1+2/(1+t))

\int_{0}^{1} (- 1 + \frac{2}{1 + t}) d t

- 1 + 2 ln (2)

Decimal Notation

0.38629 \dots

Solution steps

\int_{0}^{1} - 1 + \frac{2}{1 + t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{1} 1 d t + \int_{0}^{1} \frac{2}{1 + t} d t

\int_{0}^{1} 1 d t = 1

\int_{0}^{1} \frac{2}{1 + t} d t = 2 ln (2)

= - 1 + 2 ln (2)

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