Integral of xe^(x+1) dx

1. Rewrite the integrand:

   $$e^{x + 1} x = e x e^{x}$$

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

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

   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.

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

   So, the result is: $e \left(x e^{x} - e^{x}\right)$

3. Now simplify:

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

4. Add the constant of integration:

   $$\left(x - 1\right) e^{x + 1}+ \mathrm{constant}$$

The answer is:

$$\left(x - 1\right) e^{x + 1}+ \mathrm{constant}$$