Integral of (3x+4)*e^(3x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $du$:

      $$\int \left(\frac{u e^{u}}{3} + \frac{4 e^{u}}{3}\right)\, du$$

      1. Integrate term-by-term:

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

            $$\int \frac{u e^{u}}{3}\, du = \frac{\int u e^{u}\, du}{3}$$

            1. Use integration by parts:

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

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

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

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

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

               Now evaluate the sub-integral.

            2. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $\frac{u e^{u}}{3} - \frac{e^{u}}{3}$

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

            $$\int \frac{4 e^{u}}{3}\, du = \frac{4 \int e^{u}\, du}{3}$$

            1. The integral of the exponential function is itself.

               $$\int e^{u}\, du = e^{u}$$

            So, the result is: $\frac{4 e^{u}}{3}$

         The result is: $\frac{u e^{u}}{3} + e^{u}$

      Now substitute $u$ back in:

      $$x e^{3 x} + e^{3 x}$$

   Method #2

   1. Rewrite the integrand:

      $$e^{3 x} \left(3 x + 4\right) = 3 x e^{3 x} + 4 e^{3 x}$$

   2. Integrate term-by-term:

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

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

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

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

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

            1. Let $u = 3 x$.

               Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{3}\, du$$

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 x}}{3}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 3 x$.

               Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{3}\, du$$

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 x}}{3}$$

            So, the result is: $\frac{e^{3 x}}{9}$

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

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

         $$\int 4 e^{3 x}\, dx = 4 \int e^{3 x}\, dx$$

         1. Let $u = 3 x$.

            Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{3}\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

               So, the result is: $\frac{e^{u}}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

         So, the result is: $\frac{4 e^{3 x}}{3}$

      The result is: $x e^{3 x} + e^{3 x}$

   Method #3

   1. Rewrite the integrand:

      $$e^{3 x} \left(3 x + 4\right) = 3 x e^{3 x} + 4 e^{3 x}$$

   2. Integrate term-by-term:

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

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

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

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

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

            1. Let $u = 3 x$.

               Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{3}\, du$$

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 x}}{3}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 3 x$.

               Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

               $$\int \frac{e^{u}}{3}\, du$$

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

                  $$\text{False}$$

                  1. The integral of the exponential function is itself.

                     $$\int e^{u}\, du = e^{u}$$

                  So, the result is: $\frac{e^{u}}{3}$

               Now substitute $u$ back in:

               $$\frac{e^{3 x}}{3}$$

            So, the result is: $\frac{e^{3 x}}{9}$

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

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

         $$\int 4 e^{3 x}\, dx = 4 \int e^{3 x}\, dx$$

         1. Let $u = 3 x$.

            Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{3}\, du$$

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

               So, the result is: $\frac{e^{u}}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

         So, the result is: $\frac{4 e^{3 x}}{3}$

      The result is: $x e^{3 x} + e^{3 x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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