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

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

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

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

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

Solution steps

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

Complete the square

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

= \int \frac{x - 1}{( 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

Expand

\frac{2 u - 3}{4 u ^{2} + 3} : \frac{2 u}{4 u ^{2} + 3} - \frac{3}{4 u ^{2} + 3}

Apply the Sum Rule:

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

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

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

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

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

Substitute back

u = x + \frac{1}{2}

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

Simplify

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

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

Add a constant to the solution

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

\int 5 x + x d x

\frac{x ( x + 1 ) d y}{d x} + x y = 1, y (e) = 1

y = \frac{4 x - 7 - x}{5}

\int (cos (x))^{2} d x

\int_{0}^{1} x (x - x) d x

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