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

The integral of (x^3)/(x^4-x^2+2) is 1/4 ln|4x^4-4x^2+8|+1/(2sqrt(7))arctan((2x^2-1)/(sqrt(7)))+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^4-x^2+2)

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

\frac{1}{4} ln 4 x^{4} - 4 x^{2} + 8 + \frac{1}{2 7} arctan (\frac{2 x ^{2} - 1}{7}) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

u^{2} + 2 - u : (u - \frac{1}{2})^{2} + \frac{7}{4}

= \frac{1}{2} \cdot \int \frac{u}{( u - \frac{1}{2} ) ^{2} + \frac{7}{4}} d u

Apply u-substitution

: \int \frac{2 ( 2 v + 1 )}{4 v ^{2} + 7} d v

= \frac{1}{2} \cdot \int \frac{2 ( 2 v + 1 )}{4 v ^{2} + 7} d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{2 v + 1}{4 v ^{2} + 7} d v

Expand

\frac{2 v + 1}{4 v ^{2} + 7} : \frac{2 v}{4 v ^{2} + 7} + \frac{1}{4 v ^{2} + 7}

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 2 (\int \frac{2 v}{4 v ^{2} + 7} d v + \int \frac{1}{4 v ^{2} + 7} d v)

\int \frac{2 v}{4 v ^{2} + 7} d v = \frac{1}{4} ln 4 v^{2} + 7

\int \frac{1}{4 v ^{2} + 7} d v = \frac{1}{2 7} arctan (\frac{2}{7} v)

= \frac{1}{2} \cdot 2 (\frac{1}{4} ln 4 v^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} v))

Substitute back

= \frac{1}{2} \cdot 2 (\frac{1}{4} ln 4 (x^{2} - \frac{1}{2})^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} (x^{2} - \frac{1}{2})))

Simplify

\frac{1}{2} \cdot 2 (\frac{1}{4} ln 4 (x^{2} - \frac{1}{2})^{2} + 7 + \frac{1}{2 7} arctan (\frac{2}{7} (x^{2} - \frac{1}{2}))) : \frac{1}{4} ln 4 x^{4} - 4 x^{2} + 8 + \frac{1}{2 7} arctan (\frac{2 x ^{2} - 1}{7})

= \frac{1}{4} ln 4 x^{4} - 4 x^{2} + 8 + \frac{1}{2 7} arctan (\frac{2 x ^{2} - 1}{7})

Add a constant to the solution

= \frac{1}{4} ln 4 x^{4} - 4 x^{2} + 8 + \frac{1}{2 7} arctan (\frac{2 x ^{2} - 1}{7}) + C

f (x) = (3 x + 7) e^{x}

x^{2} + 10

\frac{d}{d x} (1.394 (1.006)^{x})

\int \frac{1}{( x ^{2} - 196 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (arccot (4 x))

What is the integral of (x^3)/(x^4-x^2+2) ?
The integral of (x^3)/(x^4-x^2+2) is 1/4 ln|4x^4-4x^2+8|+1/(2sqrt(7))arctan((2x^2-1)/(sqrt(7)))+C