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

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

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

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

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

Decimal Notation

0.11370 \dots

Solution steps

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

Expand

\frac{- x ^{2} + x}{x + 1} : - \frac{x ^{2}}{x + 1} + \frac{x}{x + 1}

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

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

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

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

Simplify

- (ln (2) - \frac{1}{2}) + 1 - ln (2) : - 2 ln (2) + \frac{3}{2}

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

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