What is the integral of 8x^3e^{4x} ?

The integral of 8x^3e^{4x} is 1/32 (64e^{4x}x^3-3(16e^{4x}x^2-2(4e^{4x}x-e^{4x})))+C

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integral of 8x^3e^{4x}

\int 8 x^{3} e^{4 x} d x

\frac{1}{32} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x}))) + C

Solution steps

\int 8 x^{3} e^{4 x} 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} e^{4 x} d x

Apply u-substitution

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

= 8 \cdot \int \frac{e ^{u} u ^{3}}{256} d u

Take the constant out:

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

= 8 \cdot \frac{1}{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

= 8 \cdot \frac{1}{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}))

= 8 \cdot \frac{1}{256} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = 4 x

= 8 \cdot \frac{1}{256} ((4 x)^{3} e^{4 x} - 3 ((4 x)^{2} e^{4 x} - 2 (e^{4 x} \cdot 4 x - e^{4 x})))

Simplify

8 \cdot \frac{1}{256} ((4 x)^{3} e^{4 x} - 3 ((4 x)^{2} e^{4 x} - 2 (e^{4 x} \cdot 4 x - e^{4 x}))) : \frac{1}{32} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))

= \frac{1}{32} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))

Add a constant to the solution

= \frac{1}{32} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x}))) + C

e^{- 2 t} t^{\frac{3}{2}}

f (x) = - 8 x^{2} - 7 x, at x = - 4

\int \frac{x ^{2} - 3 x + 12}{( x ^{2} - 4 x + 11 ) ^{2}} d x

\frac{d}{d x} (\frac{1}{x ^{3}} - \frac{2}{x ^{2}})

(x^{4} + 7 x^{2} - 3)^{4}

What is the integral of 8x^3e^{4x} ?
The integral of 8x^3e^{4x} is 1/32 (64e^{4x}x^3-3(16e^{4x}x^2-2(4e^{4x}x-e^{4x})))+C