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

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

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

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

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

Solution steps

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

Complete the square

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

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

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 ( w ^{2} + 1 )} d w

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}{w ^{2} + 1} d w

Use the common integral:

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

= 4 \cdot \frac{1}{2 3} arctan (w)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

f (x) = ln (\frac{x + 1}{x - 1})

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

5 x^{2} + 4 x + 2

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (a x)

2 x^{3} + 3 x^{2} - 12 x + 5

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