What is the integral of 8x^3(2x^4+4) ?

The integral of 8x^3(2x^4+4) is 16(1/4 x^4(x^4+2)-(x^8)/8)+C

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integral of 8x^3(2x^4+4)

\int 8 x^{3} (2 x^{4} + 4) d x

16 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8}) + C

Solution steps

\int 8 x^{3} (2 x^{4} + 4) d x

Take the constant out:

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

= 8 \cdot \int x^{3} (2 x^{4} + 4) d x

x^{3} (2 x^{4} + 4) = 2 x^{3} (x^{4} + 2)

= 8 \cdot \int 2 x^{3} (x^{4} + 2) d x

Take the constant out:

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

= 8 \cdot 2 \cdot \int x^{3} (x^{4} + 2) d x

Apply Integration By Parts

: \frac{1}{4} x^{4} (x^{4} + 2) - \int x^{7} d x

= 8 \cdot 2 (\frac{1}{4} x^{4} (x^{4} + 2) - \int x^{7} d x)

\int x^{7} d x = \frac{x ^{8}}{8}

= 8 \cdot 2 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8})

Simplify

= 16 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8})

Add a constant to the solution

= 16 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8}) + C

\frac{d}{d x} (cos (x) - \frac{1}{2})

\int_{0}^{1} x e^{3 x} d x

\int 9 x (1 - x) d x

\int_{0}^{π} - cos (x) d x

cos (x) \frac{d y}{d x} + y sin (x) = 1

What is the integral of 8x^3(2x^4+4) ?
The integral of 8x^3(2x^4+4) is 16(1/4 x^4(x^4+2)-(x^8)/8)+C