Integral of (4x-7)e^(-3x)dx dx

1. Rewrite the integrand:

   $$e^{- 3 x} \left(4 x - 7\right) = 4 x e^{- 3 x} - 7 e^{- 3 x}$$

2. Integrate term-by-term:

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

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

      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 $- \frac{du}{3}$:

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

            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:

         $$\int \left(- \frac{e^{- 3 x}}{3}\right)\, dx = - \frac{\int e^{- 3 x}\, dx}{3}$$

         1. Let $u = - 3 x$.

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

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

            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: $- \frac{4 x e^{- 3 x}}{3} - \frac{4 e^{- 3 x}}{9}$

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

      $$\int \left(- 7 e^{- 3 x}\right)\, dx = - 7 \int e^{- 3 x}\, dx$$

      1. Let $u = - 3 x$.

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

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

         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: $\frac{7 e^{- 3 x}}{3}$

   The result is: $- \frac{4 x e^{- 3 x}}{3} + \frac{17 e^{- 3 x}}{9}$

3. Now simplify:

   $$\frac{\left(17 - 12 x\right) e^{- 3 x}}{9}$$

4. Add the constant of integration:

   $$\frac{\left(17 - 12 x\right) e^{- 3 x}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(17 - 12 x\right) e^{- 3 x}}{9}+ \mathrm{constant}$$