Answer
$-\frac{1}{8}e^{-2x}(4x^3 +6x^2 +6x +3) +C$
Work Step by Step
Use the tabular method to find the indefinite integral
Let $u=x^3$, $dv=e^{-2x}dx$
Alternating Signs
+,-,+,-,+
Derivatives of u
$x^3, 3x^2, 6x, 6, 0$
Integrals of dv
$e^{-2x}, -\frac{1}{2}e^{-2x}, \frac{1}{4}e^{-2x}, -\frac{1}{8}e^{-2x} , \frac{1}{16}e^{-2x}$
Use the alternating signs, multiply the first function of u with the first integration of dv, then repeat with the next values of u and dv, alternating the signs each time
$-\frac{1}{2}x^3e^{-2x} - \frac{3}{4}x^2e^{-2x} - \frac{3}{4}xe^{-2x} - \frac{3}{8}e^{-2x} +C$
$-\frac{1}{8}e^{-2x}(4x^3 +6x^2+6x+3) +C$
