Integral of [(7+2x+8x^2)*exp(2x)] dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $du$:

      $$\int \left(u^{2} e^{u} + \frac{u e^{u}}{2} + \frac{7 e^{u}}{2}\right)\, du$$

      1. Integrate term-by-term:

         1. Use integration by parts:

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

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

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

            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. Use integration by parts:

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

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

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

            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.

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

            $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

            1. The integral of the exponential function is itself.

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

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

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

            $$\int \frac{u e^{u}}{2}\, du = \frac{\int u e^{u}\, du}{2}$$

            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: $\frac{u e^{u}}{2} - \frac{e^{u}}{2}$

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

            $$\int \frac{7 e^{u}}{2}\, du = \frac{7 \int e^{u}\, du}{2}$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $\frac{7 e^{u}}{2}$

         The result is: $u^{2} e^{u} - \frac{3 u e^{u}}{2} + 5 e^{u}$

      Now substitute $u$ back in:

      $$4 x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(8 x^{2} + \left(2 x + 7\right)\right) e^{2 x} = 8 x^{2} e^{2 x} + 2 x e^{2 x} + 7 e^{2 x}$$

   2. Integrate term-by-term:

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

         $$\int 8 x^{2} e^{2 x}\, dx = 8 \int x^{2} e^{2 x}\, dx$$

         1. Use integration by parts:

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

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            So, the result is: $\frac{e^{2 x}}{4}$

         So, the result is: $4 x^{2} e^{2 x} - 4 x e^{2 x} + 2 e^{2 x}$

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

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

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            So, the result is: $\frac{e^{2 x}}{4}$

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

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

         $$\int 7 e^{2 x}\, dx = 7 \int e^{2 x}\, dx$$

         1. Let $u = 2 x$.

            Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

            $$\int \frac{e^{u}}{2}\, du$$

            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: $\frac{e^{u}}{2}$

            Now substitute $u$ back in:

            $$\frac{e^{2 x}}{2}$$

         So, the result is: $\frac{7 e^{2 x}}{2}$

      The result is: $4 x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x}$

   Method #3

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = 8 x^{2} + 2 x + 7$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

      Then $\operatorname{du}{\left(x \right)} = 16 x + 2$.

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

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{e^{u}}{2}\, du$$

         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: $\frac{e^{u}}{2}$

         Now substitute $u$ back in:

         $$\frac{e^{2 x}}{2}$$

      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 $u{\left(x \right)} = 8 x + 1$ and let $\operatorname{dv}{\left(x \right)} = e^{2 x}$.

      Then $\operatorname{du}{\left(x \right)} = 8$.

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

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{e^{u}}{2}\, du$$

         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: $\frac{e^{u}}{2}$

         Now substitute $u$ back in:

         $$\frac{e^{2 x}}{2}$$

      Now evaluate the sub-integral.

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

      $$\int 4 e^{2 x}\, dx = 4 \int e^{2 x}\, dx$$

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{e^{u}}{2}\, du$$

         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: $\frac{e^{u}}{2}$

         Now substitute $u$ back in:

         $$\frac{e^{2 x}}{2}$$

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

   Method #4

   1. Rewrite the integrand:

      $$\left(8 x^{2} + \left(2 x + 7\right)\right) e^{2 x} = 8 x^{2} e^{2 x} + 2 x e^{2 x} + 7 e^{2 x}$$

   2. Integrate term-by-term:

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

         $$\int 8 x^{2} e^{2 x}\, dx = 8 \int x^{2} e^{2 x}\, dx$$

         1. Use integration by parts:

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

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            So, the result is: $\frac{e^{2 x}}{4}$

         So, the result is: $4 x^{2} e^{2 x} - 4 x e^{2 x} + 2 e^{2 x}$

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

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

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

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            Now evaluate the sub-integral.

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

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

            1. Let $u = 2 x$.

               Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

               $$\int \frac{e^{u}}{2}\, du$$

               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: $\frac{e^{u}}{2}$

               Now substitute $u$ back in:

               $$\frac{e^{2 x}}{2}$$

            So, the result is: $\frac{e^{2 x}}{4}$

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

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

         $$\int 7 e^{2 x}\, dx = 7 \int e^{2 x}\, dx$$

         1. Let $u = 2 x$.

            Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

            $$\int \frac{e^{u}}{2}\, du$$

            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: $\frac{e^{u}}{2}$

            Now substitute $u$ back in:

            $$\frac{e^{2 x}}{2}$$

         So, the result is: $\frac{7 e^{2 x}}{2}$

      The result is: $4 x^{2} e^{2 x} - 3 x e^{2 x} + 5 e^{2 x}$

2. Now simplify:

   $$\left(4 x^{2} - 3 x + 5\right) e^{2 x}$$

3. Add the constant of integration:

   $$\left(4 x^{2} - 3 x + 5\right) e^{2 x}+ \mathrm{constant}$$

The answer is:

$$\left(4 x^{2} - 3 x + 5\right) e^{2 x}+ \mathrm{constant}$$