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

The integral from 0 to 1 of (2-x-2/(x+1)) is -2ln(2)+3/2

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

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

- 2 ln (2) + \frac{3}{2}

Solution steps

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

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

\int_{0}^{1} 2 d x = 2

\int_{0}^{1} x d x = \frac{1}{2}

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

= 2 - \frac{1}{2} - 2 ln (2)

Combine the fractions

2 - \frac{1}{2} : \frac{3}{2}

= - 2 ln (2) + \frac{3}{2}

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\int_{e^{3}}^{e^{5}} \frac{1}{x ln ( x )} d x

\int_{0}^{x} 2 x + 1 d x

What is the integral from 0 to 1 of (2-x-2/(x+1)) ?
The integral from 0 to 1 of (2-x-2/(x+1)) is -2ln(2)+3/2