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))

integral

\frac{1}{( x ^{2} - x + 1 )}

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

Solution steps

\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

\frac{y}{( x ^{2} + y ^{2} )}

sin^{5} (x) cos^{3} (x)

\frac{x ^{2}}{( x ^{2} - 1 ) ^{2}}

\frac{x ^{2}}{( ( x - 2 ) ( x ^{2} + 1 ) )}

1 + y^{2}

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