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)