What is the integral of (x-4)/(3(x^2+x+1)) ?

The integral of (x-4)/(3(x^2+x+1)) is 1/6 (ln|4x^2+4x+4|-6sqrt(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 (x-4)/(3(x^2+x+1))

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

\frac{1}{6} (ln 4 x^{2} + 4 x + 4 - 63 arctan (\frac{2 x + 1}{3})) + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{x - 4}{x ^{2} + x + 1} : \frac{x}{x ^{2} + x + 1} - \frac{4}{x ^{2} + x + 1}

Apply the Sum Rule:

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

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

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

\int \frac{4}{x ^{2} + x + 1} d x = \frac{8}{3} arctan (\frac{2 x + 1}{3})

= \frac{1}{3} (\frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{2 x + 1}{3}) - \frac{8}{3} arctan (\frac{2 x + 1}{3}))

Simplify

\frac{1}{3} (\frac{1}{2} ln 4 x^{2} + 4 x + 4 - \frac{1}{3} arctan (\frac{2 x + 1}{3}) - \frac{8}{3} arctan (\frac{2 x + 1}{3})) : \frac{1}{6} (ln 4 x^{2} + 4 x + 4 - 63 arctan (\frac{2 x + 1}{3}))

= \frac{1}{6} (ln 4 x^{2} + 4 x + 4 - 63 arctan (\frac{2 x + 1}{3}))

Add a constant to the solution

= \frac{1}{6} (ln 4 x^{2} + 4 x + 4 - 63 arctan (\frac{2 x + 1}{3})) + C

\int \frac{8 x - 1}{2 ^{4 x^{2} - x}} d x

\int 2 x^{3} ln (6 x) d x

\int sec (t) (7 sec (t) + 5 tan (t)) d t

\int (sin (5 x))^{2} d x

\int 2^{3} d x

What is the integral of (x-4)/(3(x^2+x+1)) ?
The integral of (x-4)/(3(x^2+x+1)) is 1/6 (ln|4x^2+4x+4|-6sqrt(3)arctan((2x+1)/(sqrt(3))))+C