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.
