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

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

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

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

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

Solution steps

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

Complete the square

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

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

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} (y - \frac{1}{2}))

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} 7^{2 x} d x

x \to 0 lim (\frac{cos ( 3 x ) tan ( 3 x )}{x})

y^{^{'}} = y - 2 x

\int \frac{4 x ^{3}}{1 + x ^{4}} d x

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

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