What is the integral of 16x(8x^2+2)^2 ?

The integral of 16x(8x^2+2)^2 is 16((32x^6)/3+8x^4+2x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 16x(8x^2+2)^2

\int 16 x (8 x^{2} + 2)^{2} d x

16 (\frac{32 x ^{6}}{3} + 8 x^{4} + 2 x^{2}) + C

Solution steps

\int 16 x (8 x^{2} + 2)^{2} d x

Take the constant out:

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

= 16 \cdot \int x (8 x^{2} + 2)^{2} d x

Expand

x (8 x^{2} + 2)^{2} : 64 x^{5} + 32 x^{3} + 4 x

= 16 \cdot \int 64 x^{5} + 32 x^{3} + 4 x 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

= 16 (\int 64 x^{5} d x + \int 32 x^{3} d x + \int 4 x d x)

\int 64 x^{5} d x = \frac{32 x ^{6}}{3}

\int 32 x^{3} d x = 8 x^{4}

\int 4 x d x = 2 x^{2}

= 16 (\frac{32 x ^{6}}{3} + 8 x^{4} + 2 x^{2})

Add a constant to the solution

= 16 (\frac{32 x ^{6}}{3} + 8 x^{4} + 2 x^{2}) + C

y^{^{'}} + 8 y = 4

e^{4 x}, 0

\int 2 x^{3} - 5 x^{2} + 3 x + 1 d x

\frac{\partial}{\partial x} (ln (e^{x} + x y^{3}))

\int_{- \infty}^{0} 8^{r} d r

What is the integral of 16x(8x^2+2)^2 ?
The integral of 16x(8x^2+2)^2 is 16((32x^6)/3+8x^4+2x^2)+C