Given function #f(x)=y=ln x# with x from 1 to 3
The formula for arc length #s#
#s=int_1^3 sqrt(1+(dy/dx)^2) dx#
#dy/dx=(1/x)dx/dx=1/x#
#s=int_1^3 sqrt(1+(1/x)^2) dx#
#s=int_1^3 (sqrt(x^2+1))/x dx#
Using Algebraic Substitution
Let #z=sqrt(x^2+1)#
and #z^2=x^2+1#
Differentiating both sides
#2z *dz=2x*dx+0#
#2z *dz=2x*dx#
dividing both sides by #x^2#
#2z *dz=2x*dx#
#(2z)/x^2 *dz=(2x)/x^2*dx=(2dx)/x #
#(2z)/x^2 *dz=(2dx)/x #
#(2z*dz)/(z^2-1) =(2dx)/x #
and
#(z*dz)/(z^2-1) =(dx)/x #
Let us go back to
#s=int_1^3 (sqrt(x^2+1))/x dx=int_1^3 (sqrt(x^2+1))(dx/x)#
Replacing now with variable #z#
#s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))#
Because #z=sqrt(x^2+1)" "#
If #x# is from #x=1# to #x=3#, then, #z# is from #z=sqrt2# to #z=sqrt10#
Now we integrate
#s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))#
#s=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))#
#s=int_sqrt2^sqrt10 ((z^2*dz)/(z^2-1))#
#s=int_sqrt2^sqrt10 ((z^2-1+1)*dz)/(z^2-1)#
#s=int_sqrt2^sqrt10 ((z^2-1)/(z^2-1)+1/(z^2-1))*dz#
#s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dz#
#s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dz=[z+1/2*ln((z-1)/(z+1))]_sqrt2^sqrt10#
#s=sqrt10+1/2ln((sqrt10-1)/(sqrt10+1))-(sqrt2+1/2ln((sqrt2-1)/(sqrt2+1)))#
#color(red)(s=2.3019875345832)" "#units
God bless....I hope the explanation is useful.
