91Ó°ÊÓ

Choose the following functions: $$u = 2x$$ $$dv = e^{3x} dx$$ This choice is made because it is easy to differentiate the polynomial function, and we can easily integrate the exponential function.

Now, let's calculate the derivative du of function u and the integral v of function dv: $$du = \frac{d}{dx}(2x) = 2 dx$$ $$v = \int e^{3x} dx$$ To evaluate the integral v, we can use the substitution method: let \(w = 3x\), then \(dw = 3 dx\). Solving for \(dx\), we get \(dx = \frac{1}{3}dw\). Substituting this into the integral: $$v = \int e^w \frac{1}{3} dw = \frac{1}{3}\int e^w dw = \frac{1}{3}e^w = \frac{1}{3}e^{3x}$$

Now, we can apply the integration by parts formula: $$\int 2x e^{3x} dx = \int u dv = uv - \int v du$$ Plugging in our previously calculated values: $$\int 2x e^{3x} dx = (2x)(\frac{1}{3}e^{3x}) - \int (\frac{1}{3}e^{3x})(2 dx)$$ Simplify the expression: $$\int 2x e^{3x} dx = \frac{2}{3}xe^{3x} - 2\int \frac{1}{3}e^{3x} dx$$ Now, we can evaluate the remaining integral: $$\int \frac{1}{3}e^{3x} dx = \frac{1}{3}\int e^w \frac{1}{3} dw = \frac{1}{9}\int e^w dw = \frac{1}{9}e^w = \frac{1}{9}e^{3x}$$ Replacing this back into the expression: $$\int 2x e^{3x} dx = \frac{2}{3}xe^{3x} - 2\left(\frac{1}{9}e^{3x}\right)$$ Finally, simplify and obtain the result: $$\int 2x e^{3x} dx = \frac{2}{3}xe^{3x} - \frac{2}{9}e^{3x} + C$$ Where C is the constant of integration.