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

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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