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

The integral of (-1)/(x^2+x+1) is -2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+C

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

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

- \frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

Solution steps

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

Take the constant out:

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

= - \int \frac{1}{x ^{2} + x + 1} d x

Complete the square

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

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

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 3} d u

= - \int \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}{4 u ^{2} + 3} d u

Apply Integral Substitution

= - 4 \cdot \int \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}{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)

Substitute back

= - 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2}))

Simplify

- 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (x + \frac{1}{2})) : - \frac{2}{3} arctan (\frac{2 x + 1}{3})

= - \frac{2}{3} arctan (\frac{2 x + 1}{3})

Add a constant to the solution

= - \frac{2}{3} arctan (\frac{2 x + 1}{3}) + C

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What is the integral of (-1)/(x^2+x+1) ?
The integral of (-1)/(x^2+x+1) is -2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+C