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

The integral from 0 to 1 of (2x+3)/(x^2-x+1) is (8pi)/(3sqrt(3))

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

\int_{0}^{1} \frac{2 x + 3}{x ^{2} - x + 1} d x

\frac{8 π}{3 3}

Decimal Notation

4.83679 \dots

Solution steps

\int_{0}^{1} \frac{2 x + 3}{x ^{2} - x + 1} d x

Expand

\frac{2 x + 3}{x ^{2} - x + 1} : \frac{2 x}{x ^{2} - x + 1} + \frac{3}{x ^{2} - 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{2 x}{x ^{2} - x + 1} d x + \int_{0}^{1} \frac{3}{x ^{2} - x + 1} d x

\int_{0}^{1} \frac{2 x}{x ^{2} - x + 1} d x = \frac{2 π}{3 3}

\int_{0}^{1} \frac{3}{x ^{2} - x + 1} d x = \frac{2 π}{3}

= \frac{2 π}{3 3} + \frac{2 π}{3}

Simplify

\frac{2 π}{3 3} + \frac{2 π}{3} : \frac{8 π}{3 3}

= \frac{8 π}{3 3}

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What is the integral from 0 to 1 of (2x+3)/(x^2-x+1) ?
The integral from 0 to 1 of (2x+3)/(x^2-x+1) is (8pi)/(3sqrt(3))