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

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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`

                 `=1/2ln[(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`

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial