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`int 5x/(e^(2x)) dx` Find the indefinite integral

Recall that indefinite integral follows `int f(x) dx = F(x) +C` where:


`f(x)` as the integrand function


`F(x)` as the antiderivative of f(x)


`C` as the constant of integration.


 For the given  integral problem: `int5 x/e^(2x) dx` , we may  apply Law of exponent: `1/x^n =x^(-n)` then `1/e^(2x) = e^(-2x)` .


 `int5 x/e^(2x) dx =int5 x*e^(-2x) dx`


Apply the basic integration property: `int c*f(x) dx = c int f(x) dx` .


`int5 x*e^(-2x) dx= 5int x*e^(-2x) dx`


Apply integration by parts: `int f *g' = f*g - int g *f'du` .


Let :   `f =x` then `f' = dx`


          `g'=e^(-2x) dx` then` g = -1/2e^(-2x)`


Note:  `g = int g'= int e^(-2x) dx` . Apply u-substitution using `u =-2x` then `du = -2dx` or `(du)/(-2) =dx` .


`int e^(-2x) dx =int e^(u) * (du)/(-2)`


                      `= -1/2 int e^u du`


                      `= -1/2 e^u`


Plug-in `u =-2x` on `-1/2 e^u` , we get:   `int e^(-2x) dx =-1/2 e^(-2x)` .



Following the formula for integration by parts, we set it up as:


`5int x*e^(-2x) dx =5[ x*(-1/2 e^(-2x)) - int (-1/2 e^(-2x)) dx]`


                             `=5[ x*(-1/2 e^(-2x)) - (-1/2) int ( e^(-2x)) dx]`


                             `=(-5xe^(-2x))/2 + 5/2int ( e^(-2x)) dx`


Plug-in `int e^(-2x) dx =-1/2 e^(-2x)` , we get:


`5int x*e^(-2x) dx=(-5xe^(-2x))/2 + 5/2int ( e^(-2x)) dx`


                              `=(-5xe^(-2x))/2 + 5/2*[-1/2 e^(-2x)] +C`


                             ` =(-5xe^(-2x))/2 - (5 e^(-2x))/4 +C`


                            or   `(-5x)/(2e^(2x)) - 5/(4 e^(2x)) +C`

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