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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 ( 2 u + 1 )}{4 u ^{2} + 3} d u

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

Expand

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{π} π (sin^{2} (x))^{2} d x

\int + x ln (x) d x

e^{4 x}, e^{8 x}, x = 1

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0

\frac{d}{d x} (\frac{5}{x} + x - 4)

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