What is the integral of x^4-8x^2+16 ?

The integral of x^4-8x^2+16 is (x^5)/5-(8x^3)/3+16x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^4-8x^2+16

\int x^{4} - 8 x^{2} + 16 d x

\frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x + C

Solution steps

\int x^{4} - 8 x^{2} + 16 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 x^{4} d x - \int 8 x^{2} d x + \int 16 d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 16 d x = 16 x

= \frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{8 x ^{3}}{3} + 16 x + C

x \to - 3 lim (\frac{x}{( x + 3 ) ^{4}})

y^{^{''}} - k y = k x e^{- x}

\frac{d}{d x} (42)

y^{^{''}} + 9 y = e^{t}, y (0) = 0, y^{^{'}} (0) = 0

\int e^{(3 x + 1)} d x

What is the integral of x^4-8x^2+16 ?
The integral of x^4-8x^2+16 is (x^5)/5-(8x^3)/3+16x+C