We wish to solve int_0^(ln3) (e^(3x))/(e^(6x)+5) dx∫ln30e3xe6x+5dx.
We note that e^(6x) = (e^(3x))^2e6x=(e3x)2 and make the u-substitution u = e^(3x)u=e3x. It follows that du = 3e^(3x) dxdu=3e3xdx.
We prepare our integral for the substitution by multiplying by 3/333.
1/3 int_0^(ln3) (3e^(3x))/((e^(3x))^2 + 5) dx13∫ln303e3x(e3x)2+5dx
Note that our lower bound becomes e^(3*0) = 1e3⋅0=1 and our upper bound becomes e^(3ln3) = e^(ln(3^3)) = 3^3 = 27e3ln3=eln(33)=33=27.
1/3 int_1^(27) 1 / (u^2 + 5) du13∫2711u2+5du
This nearly looks like the proper form to yield arctan(u)arctan(u), but the 55 needs to be a 11. We factor out the 55 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) du13∫27115(u25+1)du=115∫2711(u√5)2+1du
Let v = u/sqrt(5)v=u√5. Then dv = 1/sqrt(5) dudv=1√5du. This yields the integral:
sqrt(5)/15 int_(1/sqrt(5))^(27/sqrt(5)) 1 / (v^2 + 1) dv√515∫27√51√51v2+1dv
This is the proper form to yield arctan(v)arctan(v). Evaluating yields:
sqrt(5)/15 [arctanv]_(1/sqrt(5))^(27/sqrt(5))√515[arctanv]27√51√5
= sqrt(5)/15 (arctan(27/sqrt(5)) - arctan(1/sqrt(5)))=√515(arctan(27√5)−arctan(1√5))
~~ 0.15915≈0.15915
