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

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

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

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

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

Decimal

1.19314 \dots

Solution steps

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

Expand

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

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

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

Simplify

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

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

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