#int e^x/(7+ e^2x)dx = int e^x/(7+ (e^x)^2)dx #
write it in a format that we can look to integrate more easily
Let #u = e^x#
#du = e^x dx #
The integral in terms of u is now
#int 1/(7+u^2)*du #
To use a trig substitution on this I used the fact that #int 1/(a^2+x^2)dx = 1/aarctan(x/a)#
Therefore a good trig substitution for this sort of problem would be
to let #u = sqrt(7) tan(A)# therefore #u^2 = 7tan^2(A) #
#u = sqrt(7) tan A#
#u = sqrt(7) Sin(A)/Cos(A)#
differentiating using the quotient rule
#du = sqrt(7) ((cos(A)cos(A) - sin(A)(-sin(A)))/cos^2) dA #
#Note cos^2(A) + sin ^2(A) = 1 #
#du = sqrt(7) 1/cos^2(A) dA#
The integral is now of the form
#int 1/7(1+tan^2(A)).sqrt(7) 1/cos^2(A) dA #
#= int sqrt(7)/7(1+sin^2(A)/cos^2(A)).sqrt(7) 1/cos^2(A) dA #
#= int sqrt(7)/7*1/((cos^2(A)+sin^2(A))/cos^2(A)) 1/cos^2(A) dA #
#= int sqrt(7)/7*1 dA# (everything cancels out)
#= sqrt(7)/7A#
Now remember
#u = sqrt(7) tan(A)# and #u^2 = 7tan^2(A) #
Therefore # tan(A) = u / sqrt(7)#
A =#arctan (u/sqrt(7))#
So the integral results in = #sqrt(7)/7arctan (u/sqrt(7))#
= #1/sqrt(7)arctan (u/sqrt(7))#
finally we had #u = e^x#
Therefore the answer is #1/sqrt(7) arctan (e^ x/sqrt(7))#
