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

The integral of 8x^3(2x^4+4)4 is 64(1/4 x^4(x^4+2)-(x^8)/8)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

= 32 \cdot \int x^{3} (2 x^{4} + 4) d x

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

= 32 \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

= 32 \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

= 32 \cdot 2 (\frac{1}{4} x^{4} (x^{4} + 2) - \int x^{7} d x)

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

= 32 \cdot 2 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8})

Simplify

= 64 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8})

Add a constant to the solution

= 64 (\frac{1}{4} x^{4} (x^{4} + 2) - \frac{x ^{8}}{8}) + C

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What is the integral of 8x^3(2x^4+4)4 ?
The integral of 8x^3(2x^4+4)4 is 64(1/4 x^4(x^4+2)-(x^8)/8)+C