What is the integral of (-(x^2-1)(6x^2+6))/(x^2+1) ?

The integral of (-(x^2-1)(6x^2+6))/(x^2+1) is -2x^3+6x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (-(x^2-1)(6x^2+6))/(x^2+1)

\int \frac{- ( x ^{2} - 1 ) ( 6 x ^{2} + 6 )}{x ^{2} + 1} d x

- 2 x^{3} + 6 x + C

Solution steps

\int \frac{- ( x ^{2} - 1 ) ( 6 x ^{2} + 6 )}{x ^{2} + 1} d x

Take the constant out:

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

= - \int \frac{( x ^{2} - 1 ) ( 6 x ^{2} + 6 )}{x ^{2} + 1} d x

Long division

\frac{( x ^{2} - 1 ) ( 6 x ^{2} + 6 )}{x ^{2} + 1} : 6 x^{2} - 6

= - \int 6 x^{2} - 6 d x

Apply the Sum Rule:

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

= - (\int 6 x^{2} d x - \int 6 d x)

\int 6 x^{2} d x = 2 x^{3}

\int 6 d x = 6 x

= - (2 x^{3} - 6 x)

Simplify

= - 2 x^{3} + 6 x

Add a constant to the solution

= - 2 x^{3} + 6 x + C

\int \frac{1}{0.15 x + 20} d x

f (n) = 2^{n}

\int 9 x cos (2 x) d x

a re a y = x^{2} - 1, (- 1, 1)

a re a y = x - 6, y^{2} = x

What is the integral of (-(x^2-1)(6x^2+6))/(x^2+1) ?
The integral of (-(x^2-1)(6x^2+6))/(x^2+1) is -2x^3+6x+C