Integral of (x-1)e^x dx

1. Rewrite the integrand:

   $$e^{x} \left(x - 1\right) = x e^{x} - e^{x}$$

2. Integrate term-by-term:

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

      Then $\operatorname{du}{\left(x \right)} = 1$.

      To find $v{\left(x \right)}$:

      1. The integral of the exponential function is itself.

         $$\int e^{x}\, dx = e^{x}$$

      Now evaluate the sub-integral.

   2. The integral of the exponential function is itself.

      $$\int e^{x}\, dx = e^{x}$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- e^{x}\right)\, dx = - \int e^{x}\, dx$$

      1. The integral of the exponential function is itself.

         $$\int e^{x}\, dx = e^{x}$$

      So, the result is: $- e^{x}$

   The result is: $x e^{x} - 2 e^{x}$

3. Now simplify:

   $$\left(x - 2\right) e^{x}$$

4. Add the constant of integration:

   $$\left(x - 2\right) e^{x}+ \mathrm{constant}$$

The answer is: