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

The integral of (-x^4+1)^2(-4x^3) is -1/3 (x^{12}-3x^8+3x^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (-x^4+1)^2(-4x^3)

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

- \frac{1}{3} (x^{12} - 3 x^{8} + 3 x^{4}) + C

Solution steps

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

Simplify

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

Take the constant out:

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

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

u^{\frac{1}{3}} (- u^{\frac{4}{3}} + 1)^{2} : u^{3} - 2 u^{\frac{5}{3}} + u^{\frac{1}{3}}

= - 4 \cdot \frac{1}{3} \cdot \int u^{3} - 2 u^{\frac{5}{3}} + u^{\frac{1}{3}} d u

Apply the Sum Rule:

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

= - 4 \cdot \frac{1}{3} (\int u^{3} d u - \int 2 u^{\frac{5}{3}} d u + \int u^{\frac{1}{3}} d u)

\int u^{3} d u = \frac{u ^{4}}{4}

\int 2 u^{\frac{5}{3}} d u = \frac{3}{4} u^{\frac{8}{3}}

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

= - 4 \cdot \frac{1}{3} (\frac{u ^{4}}{4} - \frac{3}{4} u^{\frac{8}{3}} + \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = x^{3}

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

Simplify

- 4 \cdot \frac{1}{3} (\frac{( x ^{3} ) ^{4}}{4} - \frac{3}{4} (x^{3})^{\frac{8}{3}} + \frac{3}{4} (x^{3})^{\frac{4}{3}}) : - \frac{1}{3} (x^{12} - 3 x^{8} + 3 x^{4})

= - \frac{1}{3} (x^{12} - 3 x^{8} + 3 x^{4})

Add a constant to the solution

= - \frac{1}{3} (x^{12} - 3 x^{8} + 3 x^{4}) + C

\frac{d}{d x} (\frac{8 x ^{3} - 6}{x ^{2}})

y^{^{'}} = - x

d er i v a t i v e y = (\frac{x ^{2} + 6}{x ^{2} - 6})^{3}

\frac{d}{d x} (arcsin (3 x) + arccos (3 x))

\frac{d}{d x} (arccot (\frac{x}{8}))

What is the integral of (-x^4+1)^2(-4x^3) ?
The integral of (-x^4+1)^2(-4x^3) is -1/3 (x^{12}-3x^8+3x^4)+C