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

The integral of 1/(sqrt(x^2-x+1)) is ln| 2/(sqrt(3))x-1/(sqrt(3))+sqrt(1/3 (4x^2-4x+4))|+C

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

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

ln \frac{2}{3} x - \frac{1}{3} + \frac{1}{3} (4 x^{2} - 4 x + 4) + 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{2}{4 u ^{2} + 3} d u

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{2} d v

= 2 \cdot \int \frac{sec ( v )}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2})))

Simplify

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

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

Add a constant to the solution

= ln \frac{2}{3} x - \frac{1}{3} + \frac{1}{3} (4 x^{2} - 4 x + 4) + C

f (x) = ln (x + x^{2} + 1)

\int 49 - t^{2} d t

\frac{\partial}{\partial x} (e^{- 4 x} cos (5 x))

\frac{\partial}{\partial x} (\frac{( x - 3 y )}{x + 3 y})

What is the integral of 1/(sqrt(x^2-x+1)) ?
The integral of 1/(sqrt(x^2-x+1)) is ln| 2/(sqrt(3))x-1/(sqrt(3))+sqrt(1/3 (4x^2-4x+4))|+C