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

The integral of (x^3)/(x^2+x+1) is 1/2 x^2-x+2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+1/8-3/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(x^2+x+1)

\int \frac{x ^{3}}{x ^{2} + x + 1} d x

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

Solution steps

\int \frac{x ^{3}}{x ^{2} + x + 1} d x

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

v^{2} + 2 v + 4 : (v + 1)^{2} + 3

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

Apply u-substitution

: \int \frac{( w - 1 ) ^{3}}{w ^{2} + 3} d w

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{( w - 1 ) ^{3}}{w ^{2} + 3} d w

Long division

\frac{( w - 1 ) ^{3}}{w ^{2} + 3} : w - 3 + \frac{8}{w ^{2} + 3}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int w - 3 + \frac{8}{w ^{2} + 3} d w

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}{2} \cdot \frac{1}{2} (\int w d w - \int 3 d w + \int \frac{8}{w ^{2} + 3} d w)

\int w d w = \frac{w ^{2}}{2}

\int 3 d w = 3 w

\int \frac{8}{w ^{2} + 3} d w = \frac{8}{3} arctan (\frac{w}{3})

= \frac{1}{2} \cdot \frac{1}{2} (\frac{w ^{2}}{2} - 3 w + \frac{8}{3} arctan (\frac{w}{3}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x^{2} y^{^{''}} - 7 x y^{^{'}} - 20 y = 0

y^{^{''}} + 9 y = 30 sin (t)

f (x) = e^{x} x

y = 8 cos (3 x), y = 8 sin (3 x), (0, \frac{π}{12})

f (x) = \frac{5}{x}, (1, 5)

What is the integral of (x^3)/(x^2+x+1) ?
The integral of (x^3)/(x^2+x+1) is 1/2 x^2-x+2/(sqrt(3))arctan((2x+1)/(sqrt(3)))+1/8-3/4+C