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

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

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

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

\frac{π}{3}

Decimal Notation

1.81379 \dots

Solution steps

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

Complete the square

x^{2} + x + 1 : (x + \frac{1}{2})^{2} + \frac{3}{4}

= \int_{- 1}^{1} \frac{1}{( x + \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

Apply u-substitution

: \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 u ^{2} + 3} d u

= \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{4}{4 u ^{2} + 3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{1}{4 u ^{2} + 3} d u

Apply Integral Substitution

= 4 \cdot \int_{- \frac{1}{3}}^{3} \frac{1}{2 3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{2 3} \cdot \int_{- \frac{1}{3}}^{3} \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot \frac{1}{2 3} [arctan (v)]_{- \frac{1}{3}}^{3}

Simplify

4 \cdot \frac{1}{2 3} [arctan (v)]_{- \frac{1}{3}}^{3} : \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{3}

= \frac{2}{3} [arctan (v)]_{- \frac{1}{3}}^{3}

Compute the boundaries:

\frac{π}{2}

= \frac{2}{3} \cdot \frac{π}{2}

Simplify

= \frac{π}{3}

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