Integral of (3x+1)e^x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\left(3 x + 1\right) e^{x} = 3 x e^{x} + e^{x}$$

   2. Integrate term-by-term:

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

         $$\int 3 x e^{x}\, dx = 3 \int x e^{x}\, dx$$

         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: $3 x e^{x} - 3 e^{x}$

      1. The integral of the exponential function is itself.

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

      The result is: $3 x e^{x} - 2 e^{x}$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = 3 x + 1$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

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

      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 a constant times a function is the constant times the integral of the function:

      $$\int 3 e^{x}\, dx = 3 \int e^{x}\, dx$$

      1. The integral of the exponential function is itself.

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

      So, the result is: $3 e^{x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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