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

      $$\left(3 x + 4\right) e^{3 x} 1 = 3 x e^{3 x} + 4 e^{3 x}$$

   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. There are multiple ways to do this integral.

               Method #1

               1. Let $u = 3 x$.

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

                  $$\int \frac{e^{u}}{9}\, du$$

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

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

                     1. The integral of the exponential function is itself.

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

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

                  Now substitute $u$ back in:

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

               Method #2

               1. Let $u = e^{3 x}$.

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

                  $$\int \frac{1}{9}\, du$$

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

                     $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

                     1. The integral of a constant is the constant times the variable of integration:

                        $$\int 1\, du = u$$

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

                  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 $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

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

                  1. The integral of the exponential function is itself.

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

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

               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 $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

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

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

               1. The integral of the exponential function is itself.

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

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

            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. Use integration by parts:

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

      Let $u{\left(x \right)} = 3 x + 4$ and let $\operatorname{dv}{\left(x \right)} = e^{3 x}$.

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

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

      1. Let $u = 3 x$.

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

         $$\int \frac{e^{u}}{9}\, du$$

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

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

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

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

      Now evaluate the sub-integral.

   2. Let $u = 3 x$.

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

      $$\int \frac{e^{u}}{9}\, du$$

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

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

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

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

   Method #4

   1. Rewrite the integrand:

      $$\left(3 x + 4\right) e^{3 x} 1 = 3 x e^{3 x} + 4 e^{3 x}$$

   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 $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

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

                  1. The integral of the exponential function is itself.

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

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

               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 $\frac{du}{3}$:

               $$\int \frac{e^{u}}{9}\, du$$

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

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

                  1. The integral of the exponential function is itself.

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

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

               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 $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

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

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

               1. The integral of the exponential function is itself.

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

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

            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:

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