#f(x)=e^(-3x)#
#f"'"(x)=-3e^(-3x)#
#L=int_a^b sqrt(1+[f"'"(x)]^2).dx#
#F(x)=int sqrt(1+9e^(-6x)).dx#
#F(x)=int sqrt(e^(6x)+9)*e^(-3x)*dx#
Making
#e^(3x)=3tany#
#cancel3*e^(3x)*dx=cancel3sec^2y*dy#
#(e^(-3x))^2*e^(3x)*dx=(sec^2y*dy)/(3tany)^2#
#e^(-3x)*dx=(sec^2y*dy)/(9tan^2y)#
So
#F(y)=int 3secy*(sec^2y*dy)/(9tan^2y)=1/3int(1/cos^3y)*(cos^2y/sin^2y)*dy#
#F(y)=1/3int dy/(cosy.sin^2y)#
But
#1/(cosy.sin^2y)=1/cosy+cosy/sin^2y#
Then
#F(y)=1/3int secy*dy+1/3intcosy/sin^2y*dy#
Solving the last term of the expression above
#1/3intcosy/sin^2y*dy#
Making
#siny=u# => #cosy*dy=du#
Resulting in
#=1/3int (du)/u^2=-1/3*1/u=-1/3*1/siny#
Back to the main expression
#F(y)=1/3*ln|secy+tany|-1/3*1/siny#
But
#tany=e^(3x)/3# => #siny=e^(3x)/3*cosy# => #(e^(6x)/9+1)cos^2y=1# => #cosy=3/sqrt(e^(6x)+9)#
#-> siny=e^(3x)/sqrt(e^(6x)+9)#
Therefore
#F(x)=1/3*ln|(sqrt(e^(6x)+9)+e^(3x))/3|-1/3*sqrt(e^(6x)+9)/e^(3x) + const.#
Finally
#L=F(x=2)-F(x=1)#
#L=1/3*ln((sqrt(e^12+9)+e^6)/3)-1/3*sqrt(e^12+9)/e^6-[1/3*ln((sqrt(e^6+9)+e^3)/3)-1/3*sqrt(e^6+9)/e^3]#
#L=1/3*ln((sqrt(e^12+9)+e^6)/(sqrt(e^6+9)+e^3))+1/3(sqrt(e^6+9)/e^3-sqrt(e^12+9)/e^6)#
