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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

- \frac{3 ( z ^{2} + 1 )}{z ^{2}}

y^{2} = 6 - x, 3 y = x + 12

\int_{π}^{2 π} 9 x sin (x) d x

n = 0 \sum \infty 12 \cdot 626582

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

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