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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} + x + 1} d x

Complete the square

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

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

Apply u-substitution

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

= 2 \cdot \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

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

Apply Integral Substitution

= 2 \cdot 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

= 2 \cdot 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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 9 - t^{2} d t

x + 2, x^{5} + 2

\frac{d}{d x} (- 2 sec^{2} (x))

x^{2} d y = (y^{2} - x y + x^{2}) d x

x + \frac{6}{x}

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