Integrating $\int xe^{-x} dx$ without parts Can $\int xe^{-x} dx$ ever be solved by integration by substitution without using parts. Or does, as I suspect, substitution fail to yield a solution in this case.
Seems that we can't get a reciprocal to cancel $x$ out under any circumstance. I know how to solve with parts, please don't use this.
 A: Differentiation under the integral sign.
The integral is precisely $$-\frac{1}{a} \dfrac{d}{da} \int e^{-a x} \ dx$$
evaluated at $a=1$.
But that is $$-\dfrac{1}{a} \dfrac{d}{da} \left(-\frac{1}{a} e^{-ax} \right) = -\frac{1}{a} \left( \frac{e^{-ax}}{a^2} + \frac{e^{-a x}x}{a} \right)$$
Evaluating at $1$ yields $$-(e^{-x} + x e^{-x}) = -e^{-x}(1+x)$$
A: By method of undetermined coefficients
Guessing the anti-derivative $F(x)$ is in the form of $(A+Bx)e^{-x}$, then $F'(x)=-Ae^{-x}+Be^{-x}-Bxe^{-x}$ implying $B=-1$ and $B-A=0$.
A: Consider a basis $\mathcal{B} = \{e^{-x}, xe^{-x}\}$. Differentiating each element of the basis we have
\begin{align*}
\frac{d}{dx}(e^{-x}) &= -e^{-x}\\
\frac{d}{dx}(xe^{-x}) &=e^{-x} + -xe^{-x}
\end{align*}
The matrix representation of the derivative operator with respect to $\mathcal{B}$ is
$$T = \begin{bmatrix}
-1 & 1\\
0 & -1
\end{bmatrix}$$
and 
$$T^{-1} = \begin{bmatrix}
-1 -1\\
0 -1
\end{bmatrix}.$$
So
$$\int xe^{-x} \operatorname{d}\!x = T^{-1}\begin{bmatrix}
0\\
1
\end{bmatrix} = -e^{-x} - xe^{-x}.$$
A: Since a solution by substitution is required, substitute $y=x$, then proceed as in one of the other answers that involve neither parts, nor substitution.
A: Let's write $$e^{-x}=\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}$$ then 
\begin{align}
\int x e^{-x}dx&=\int \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n!}dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\int x^{n+1}dx\\
&=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+2}}{n!(n+2)}+k\\
&=\sum_{n=0}^{\infty}\frac{(n+1)(-1)^{n+2}x^{n+2}}{(n+2)!}+k\\
&=-\sum_{n=0}^{\infty}\frac{n(-1)^{n}x^{n+1}}{(n+1)!}+k\\
&=-\sum_{n=0}^{\infty}\big(\frac{1}{n!}-\frac{1}{(n+1)!}\big)(-1)^{n}x^{n+1}+k\\
&=-\sum_{n=0}^{\infty}\big(\frac{1}{n!}\big)(-1)^{n}x^{n+1}+\sum_{n=0}^{\infty}\big(\frac{1}{(n+1)!}\big)(-1)^{n}x^{n+1}+k\\
&=-xe^{-x}-e^{-x}+1+k\\
&=-xe^{-x}-e^{-x}+k'\\
\end{align}
