Answer
-$e^{-x}$(3$x^2$+5x+7)+C
Work Step by Step
-- If we apply tabular integration by parts to this problem, $\int$p(x)f(x)dx, the polynomial p(x)=3$x^2$-x+2 & f(x)=$e^{-x}$. Let’s set up a table showing the $n^{th}$ term of the result of the integration (shown in the attached image below).
-- From the table, we write down the result of the integration:
$\Sigma$($\pm$)[$p^{(n-1)}$(x)][$I^{n}$f(x)]+C]
Then simplify it if necessary:
(3$x^2$-x+2)(-$e^{-x}$)-(6x-1)$e^{-x}$+6 (-$e^{-x}$)+C
When we factor out $e^{-x}$, the expression=$e^{-x}$[-(3$x^2$-x+2)-(6x-1)-6]+C
Finally simplify it, = -$e^{-x}$(3$x^2$+5x+7)+C
