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

The integral of x^4e^{3x} is 1/3 e^{3x}x^4-4/81 e^{3x}(9x^3-9x^2+6x-2)+C

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integral of x^4e^{3x}

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

\frac{1}{3} e^{3 x} x^{4} - \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{3} e^{3 x} x^{4} - \int \frac{4}{3} e^{3 x} x^{3} d x

= \frac{1}{3} e^{3 x} x^{4} - \int \frac{4}{3} e^{3 x} x^{3} d x

\int \frac{4}{3} e^{3 x} x^{3} d x = \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

= \frac{1}{3} e^{3 x} x^{4} - \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

Add a constant to the solution

= \frac{1}{3} e^{3 x} x^{4} - \frac{4}{81} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

\frac{d}{d x} (2 sec (2 x) tan (2 x))

0.125 x 64 - x^{2}

\frac{\partial}{\partial x} (x^{2} + sin (x y))

\frac{d}{d x} (a cos (b x))

\int 4 x - \frac{4}{x} d x

What is the integral of x^4e^{3x} ?
The integral of x^4e^{3x} is 1/3 e^{3x}x^4-4/81 e^{3x}(9x^3-9x^2+6x-2)+C