We wish to solve #int_0^(ln3) (e^(3x))/(e^(6x)+5) dx#.
We note that #e^(6x) = (e^(3x))^2# and make the u-substitution #u = e^(3x)#. It follows that #du = 3e^(3x) dx#.
We prepare our integral for the substitution by multiplying by #3/3#.
#1/3 int_0^(ln3) (3e^(3x))/((e^(3x))^2 + 5) dx#
Note that our lower bound becomes #e^(3*0) = 1# and our upper bound becomes #e^(3ln3) = e^(ln(3^3)) = 3^3 = 27#.
#1/3 int_1^(27) 1 / (u^2 + 5) du#
This nearly looks like the proper form to yield #arctan(u)#, but the #5# needs to be a #1#. We factor out the #5# and make another substitution.
#1/3 int_1^(27) 1 / (5(u^2/5 + 1))du = 1/15 int_1^(27) 1 / ((u/sqrt(5))^2 + 1) du#
Let #v = u/sqrt(5)#. Then #dv = 1/sqrt(5) du#. This yields the integral:
#sqrt(5)/15 int_(1/sqrt(5))^(27/sqrt(5)) 1 / (v^2 + 1) dv#
This is the proper form to yield #arctan(v)#. Evaluating yields:
#sqrt(5)/15 [arctanv]_(1/sqrt(5))^(27/sqrt(5))#
# = sqrt(5)/15 (arctan(27/sqrt(5)) - arctan(1/sqrt(5)))#
#~~ 0.15915#
