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

The integral of e^{3x}x^3 is 1/27 e^{3x}(9x^3-9x^2+6x-2)+C

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integral of e^{3x}x^3

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

\frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

Solution steps

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

Apply u-substitution

= \int \frac{e ^{u} u ^{3}}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

= \frac{1}{81} (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}{81} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = 3 x

= \frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})))

Simplify

\frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x}))) : \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

= \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

Add a constant to the solution

= \frac{1}{27} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

What is the integral of e^{3x}x^3 ?
The integral of e^{3x}x^3 is 1/27 e^{3x}(9x^3-9x^2+6x-2)+C