What is the integral of 4x(2x^2+1)8 ?

The integral of 4x(2x^2+1)8 is 32((x^4)/2+(x^2)/2)+C

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integral of 4x(2x^2+1)8

\int 4 x (2 x^{2} + 1) 8 d x

32 (\frac{x ^{4}}{2} + \frac{x ^{2}}{2}) + C

Solution steps

\int 4 x (2 x^{2} + 1) \cdot 8 d x

Take the constant out:

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

= 4 \cdot 8 \cdot \int x (2 x^{2} + 1) d x

Simplify

= 32 \cdot \int x (2 x^{2} + 1) d x

Expand

x (2 x^{2} + 1) : 2 x^{3} + x

= 32 \cdot \int 2 x^{3} + 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

= 32 (\int 2 x^{3} d x + \int x d x)

\int 2 x^{3} d x = \frac{x ^{4}}{2}

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

= 32 (\frac{x ^{4}}{2} + \frac{x ^{2}}{2})

Add a constant to the solution

= 32 (\frac{x ^{4}}{2} + \frac{x ^{2}}{2}) + C

\frac{d y}{d t} - (\frac{3}{t}) \cdot y = 2 t^{3} \cdot e^{(2 t)}

\frac{d}{d t} (5 e^{- 2 t} + 3 t e^{- 2 t} - 2)

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (1) = 0, y^{^{'}} (1) = 4

x cos (x)

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

What is the integral of 4x(2x^2+1)8 ?
The integral of 4x(2x^2+1)8 is 32((x^4)/2+(x^2)/2)+C