`int 2 / (7e^x + 4) dx` Find the indefinite integral

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To evaluate the given integral problem:` int 2/(7e^x+4)dx` , we may apply u-substitution using:  `u= e^x ` then `du = e^x dx` .

 Plug-in `u = e^x` on `du= e^x dx` , we get: `du = u dx` or `(du)/u =dx`

The integral becomes:

`int 2/(7e^x+4)dx =int 2/(7u+4)* (du)/u`

                         `=int 2/(7u^2+4u)du`

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

`int 2/(7u^2+4u)du =2int 1/(7u^2+4u)du`

Apply completing the square: `7u^2+4u =(sqrt(7)u+2/sqrt(7))^2 -4/7`

`2int 1/(7u^2+4u)du =2int 1/((sqrt(7)u+2/sqrt(7))^2 -4/7)du`



Let `v =sqrt(7)u+2/sqrt(7)` then `dv = sqrt(7) du`  or `(dv)/sqrt(7) = du` .

The integral becomes: 

`2int 1/(7u^2+4u)du =2 int 1/(v^2 -4/7) *(dv)/sqrt(7)`

Rationalize the denominator:

`2 int 1/(v^2 -4/7) *(dv)/sqrt(7) *sqrt(7)/sqrt(7)`

`= 2 int (sqrt(7)dv)/ ( 7*(v^2 -4/7))`

`=2 int (sqrt(7)dv)/ ( 7v^2 -4)`


From the table of integrals, we may apply `int dx/(x^2-a^2) = 1/(2a)ln[(u-a)/(u+a)]+C`

 Let:` w = sqrt(7)v` then `dw = sqrt(7) dv`

`2int (sqrt(7) dv)/ ( 7v^2 -4) =2int (sqrt(7) dv)/ (( sqrt(7)v)^2 -2^2)`

                  `= 2 int (dw)/ (w^2-2^2)`

                 `= 2 *1/(2*2)ln[(w-2)/(w+2)]+C`


Recall we let: `w =sqrt(7)v` and `v =sqrt(7)u+2/sqrt(7)` .

Then, `w=sqrt(7)*[sqrt(7)u+2/sqrt(7)] = 7u +2`

Plug-in `u =e^x` on `w=7u +2` , we get: `w= 7e^x+2`

The indefinite integral will be:

`int 2/(7e^x+4)dx =1/2ln[(7e^x+2-2)/(7e^x+2+2)]+C`

                     ` =1/2ln[(7e^x)/(7e^x+4)]+C`



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