I assume the integral is
[tex]\displaystyle \int x^3 e^{-x} \, dx[/tex]
Integrate by parts with
[tex]u = x^3 \implies du = 3x^2 \, dx[/tex]
[tex]dv = e^{-x} \, dx \implies v = -e^{-x}[/tex]
so that
[tex]\displaystyle \int u \, dv = uv - \int v \, du[/tex]
i.e.
[tex]\displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3 \int x^2 e^{-x} \, dx[/tex]
Integrate by parts again, this time with
[tex]u = x^2 \implies du = 2x \, dx[/tex]
[tex]dv = e^{-x} \, dx \implies v = -e^{-x}[/tex]
and so
[tex]\displaystyle \int x^2 e^{-x} \, dx = -x^2 e^{-x} + 2 \int x e^{-x} \, dx[/tex]
[tex]\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} + 3 \left(-x^2 e^{-x} + 2 \int x e^{-x} \, dx\right)[/tex]
[tex]\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} + 6 \int x e^{-x} \, dx[/tex]
Once more, with
[tex]u = x \implies du = dx[/tex]
[tex]dv = e^{-x} \, dx \implies v = -e^{-x}[/tex]
[tex]\displaystyle \int x e^{-x} \, dx = -xe^{-x} + \int e^{-x} \, dx[/tex]
[tex]\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} + 6 \left(-xe^{-x} + \int e^{-x} \, dx\right)[/tex]
[tex]\implies \displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} -6xe^{-x} + 6 \int e^{-x} \, dx[/tex]
Finally,
[tex]\displaystyle \int e^{-x} \, dx = -e^{-x} + C[/tex]
and so
[tex]\displaystyle \int x^3 e^{-x} \, dx = -x^3 e^{-x} - 3x^2 e^{-x} -6xe^{-x} - 6 e^{-x} + C[/tex]
or equivalently,
[tex]\displaystyle \int x^3 e^{-x} \, dx = \boxed{-(x^3+3x^2+6x+6) e^{-x} + C}[/tex]
We can also approach the integral more generally by considering
[tex]I_n = \displaystyle \int x^n e^{-x} \, dx[/tex]
Integrate by parts with
[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]
[tex]dv = e^{-x} \, dx \implies v = -e^{-x}[/tex]
Then
[tex]\displaystyle I_n = -x^n e^{-x} + n \int x^{n-1} e^{-x} \, dx = -x^n e^{-x} + n I_{n-1}[/tex]
We can solve the recurrence using the initial "value" [tex]I_0 = -e^{-x}+C[/tex] to find a general formula for [tex]I_n[/tex]. Then we get the integral we want [tex]I_3[/tex] "for free".
