Integral of (x^3)(e^x) dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

   Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$.

   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. Use integration by parts:

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

   Let $u{\left(x \right)} = 3 x^{2}$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

   Then $\operatorname{du}{\left(x \right)} = 6 x$.

   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.

3. Use integration by parts:

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

   Let $u{\left(x \right)} = 6 x$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

   Then $\operatorname{du}{\left(x \right)} = 6$.

   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.

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

   $$\int 6 e^{x}\, dx = 6 \int e^{x}\, dx$$

   1. The integral of the exponential function is itself.

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

   So, the result is: $6 e^{x}$

5. Now simplify:

   $$\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}$$

6. Add the constant of integration:

   $$\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}+ \mathrm{constant}$$

The answer is:

$$\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}+ \mathrm{constant}$$