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

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