What is the integral of x^3e^{6x} ?

The integral of x^3e^{6x} is 1/216 e^{6x}(36x^3-18x^2+6x-1)+C

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integral of x^3e^{6x}

\int x^{3} e^{6 x} d x

\frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) + C

Solution steps

\int x^{3} e^{6 x} d x

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{1296} d u

= \int \frac{e ^{u} u ^{3}}{1296} d u

Take the constant out:

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

= \frac{1}{1296} \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}{1296} (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}{1296} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = 6 x

= \frac{1}{1296} ((6 x)^{3} e^{6 x} - 3 ((6 x)^{2} e^{6 x} - 2 (e^{6 x} \cdot 6 x - e^{6 x})))

Simplify

\frac{1}{1296} ((6 x)^{3} e^{6 x} - 3 ((6 x)^{2} e^{6 x} - 2 (e^{6 x} \cdot 6 x - e^{6 x}))) : \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)

= \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)

Add a constant to the solution

= \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) + C

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What is the integral of x^3e^{6x} ?
The integral of x^3e^{6x} is 1/216 e^{6x}(36x^3-18x^2+6x-1)+C