What is the integral of 4x^3(2x^4-e) ?

The integral of 4x^3(2x^4-e) is x^8-ex^4+C

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integral of 4x^3(2x^4-e)

\int 4 x^{3} (2 x^{4} - e) d x

x^{8} - e x^{4} + C

Solution steps

\int 4 x^{3} (2 x^{4} - e) d x

Take the constant out:

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

= 4 \cdot \int x^{3} (2 x^{4} - e) d x

Expand

x^{3} (2 x^{4} - e) : 2 x^{7} - e x^{3}

= 4 \cdot \int 2 x^{7} - e x^{3} 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

= 4 (\int 2 x^{7} d x - \int e x^{3} d x)

\int 2 x^{7} d x = \frac{x ^{8}}{4}

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

= 4 (\frac{x ^{8}}{4} - \frac{e x ^{4}}{4})

Simplify

4 (\frac{x ^{8}}{4} - \frac{e x ^{4}}{4}) : x^{8} - e x^{4}

= x^{8} - e x^{4}

Add a constant to the solution

= x^{8} - e x^{4} + C

\int \frac{( x ^{m} - x ^{n} ) ^{2}}{x} d x

\int_{0}^{3} 2 π (x + 7) (2 x) d x

y^{^{''}} + y = cos (x)

f (x) = x^{3} - 27 x

\frac{d}{d x} (x^{x l n (x)})

What is the integral of 4x^3(2x^4-e) ?
The integral of 4x^3(2x^4-e) is x^8-ex^4+C