Integral of e^(x)*(x+n)^(2) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\left(n + x\right)^{2} e^{x} = n^{2} e^{x} + 2 n x e^{x} + x^{2} e^{x}$$

   2. Integrate term-by-term:

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

         $$\int n^{2} e^{x}\, dx = n^{2} \int e^{x}\, dx$$

         1. The integral of the exponential function is itself.

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

         So, the result is: $n^{2} e^{x}$

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

         $$\int 2 n x e^{x}\, dx = 2 n \int x e^{x}\, dx$$

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

      1. Use integration by parts:

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

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

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

         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.

      2. Use integration by parts:

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

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

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

         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.

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

         $$\int 2 e^{x}\, dx = 2 \int e^{x}\, dx$$

         1. The integral of the exponential function is itself.

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

         So, the result is: $2 e^{x}$

      The result is: $n^{2} e^{x} + 2 n \left(x e^{x} - e^{x}\right) + x^{2} e^{x} - 2 x e^{x} + 2 e^{x}$

   Method #2

   1. Use integration by parts:

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

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

      Then $\operatorname{du}{\left(x \right)} = 2 n + 2 x$.

      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.

   2. Use integration by parts:

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

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

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

      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.

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

      $$\int 2 e^{x}\, dx = 2 \int e^{x}\, dx$$

      1. The integral of the exponential function is itself.

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

      So, the result is: $2 e^{x}$

2. Now simplify:

   $$\left(n^{2} + 2 n \left(x - 1\right) + x^{2} - 2 x + 2\right) e^{x}$$

3. Add the constant of integration:

   $$\left(n^{2} + 2 n \left(x - 1\right) + x^{2} - 2 x + 2\right) e^{x}+ \mathrm{constant}$$

The answer is:

$$\left(n^{2} + 2 n \left(x - 1\right) + x^{2} - 2 x + 2\right) e^{x}+ \mathrm{constant}$$