What is the integral of y=x^3(e^x)^4 ?

The integral of y=x^3(e^x)^4 is 1/256 (64e^{4x}x^3-3(16e^{4x}x^2-2(4e^{4x}x-e^{4x})))+C

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integral of y=x^3(e^x)^4

integral

y = x^{3} (e^{x})^{4}

\frac{1}{256} (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 x^{3} (e^{x})^{4} d x

(e^{x})^{4} = e^{4 x}

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

Apply u-substitution

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

= \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

= \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

= \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}))

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

Substitute back

u = 4 x

= \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

\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}{256} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x})))

= \frac{1}{256} (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}{256} (64 e^{4 x} x^{3} - 3 (16 e^{4 x} x^{2} - 2 (4 e^{4 x} x - e^{4 x}))) + C

\frac{1}{3 x ^{2} - 3 x}

8 y^{3}

cos (y^{3})

\frac{( e ^{t} ) ^{2}}{( e ^{2 t} + 1 )}

\frac{x ^{2} + 1}{x ^{2} - 1}

What is the integral of y=x^3(e^x)^4 ?
The integral of y=x^3(e^x)^4 is 1/256 (64e^{4x}x^3-3(16e^{4x}x^2-2(4e^{4x}x-e^{4x})))+C