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

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

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

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

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

Solution steps

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

Complete the square

1 + x + x^{2} : (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{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 ( v ^{2} + 1 )} d v

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}{v ^{2} + 1} d v

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int y^{- 3} d y

\frac{\partial}{\partial y} (x y^{2} arctan (z))

(4 x + 2 y) d x + (2 x + 2 y) d y = 0

x^{2} + y^{2} = 1, y = 1 - x

n = 1 \sum \infty \frac{3}{n ^{3}}

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