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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

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

= 2 \cdot \frac{1}{2} (- \frac{1}{2} ln v^{2} + 1 + 3 arctan (v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

4 x^{3} - 5 7 x^{6} + π x

e^{- 4 t} cos (3 t)

\int x e^{3 x} d x

f (t) = 6 cos (t)

\frac{d}{d x} (18 x (x^{2} + 6)^{8})

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