What is the integral of 15x^3e^{x/4} ?

The integral of 15x^3e^{x/4} is 3840(1/64 e^{x/4}x^3-3(1/16 e^{x/4}x^2-2(1/4 e^{x/4}x-e^{x/4})))+C

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integral of 15x^3e^{x/4}

\int 15 x^{3} e^{\frac{x}{4}} d x

3840 (\frac{1}{64} e^{\frac{x}{4}} x^{3} - 3 (\frac{1}{16} e^{\frac{x}{4}} x^{2} - 2 (\frac{1}{4} e^{\frac{x}{4}} x - e^{\frac{x}{4}}))) + C

Solution steps

\int 15 x^{3} e^{\frac{x}{4}} d x

Take the constant out:

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

= 15 \cdot \int x^{3} e^{\frac{x}{4}} d x

Apply u-substitution

: \int 256 e^{u} u^{3} d u

= 15 \cdot \int 256 e^{u} u^{3} d u

Take the constant out:

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

= 15 \cdot 256 \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= 15 \cdot 256 (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= 15 \cdot 256 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = \frac{x}{4}

= 15 \cdot 256 ((\frac{x}{4})^{3} e^{\frac{x}{4}} - 3 ((\frac{x}{4})^{2} e^{\frac{x}{4}} - 2 (e^{\frac{x}{4}} \frac{x}{4} - e^{\frac{x}{4}})))

Simplify

15 \cdot 256 ((\frac{x}{4})^{3} e^{\frac{x}{4}} - 3 ((\frac{x}{4})^{2} e^{\frac{x}{4}} - 2 (e^{\frac{x}{4}} \frac{x}{4} - e^{\frac{x}{4}}))) : 3840 (\frac{1}{64} e^{\frac{x}{4}} x^{3} - 3 (\frac{1}{16} e^{\frac{x}{4}} x^{2} - 2 (\frac{1}{4} e^{\frac{x}{4}} x - e^{\frac{x}{4}})))

= 3840 (\frac{1}{64} e^{\frac{x}{4}} x^{3} - 3 (\frac{1}{16} e^{\frac{x}{4}} x^{2} - 2 (\frac{1}{4} e^{\frac{x}{4}} x - e^{\frac{x}{4}})))

Add a constant to the solution

= 3840 (\frac{1}{64} e^{\frac{x}{4}} x^{3} - 3 (\frac{1}{16} e^{\frac{x}{4}} x^{2} - 2 (\frac{1}{4} e^{\frac{x}{4}} x - e^{\frac{x}{4}}))) + C

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What is the integral of 15x^3e^{x/4} ?
The integral of 15x^3e^{x/4} is 3840(1/64 e^{x/4}x^3-3(1/16 e^{x/4}x^2-2(1/4 e^{x/4}x-e^{x/4})))+C