Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 34

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