Given function f(x)=y=ln xf(x)=y=lnx with x from 1 to 3
The formula for arc length ss
s=int_1^3 sqrt(1+(dy/dx)^2) dxs=∫31√1+(dydx)2dx
dy/dx=(1/x)dx/dx=1/xdydx=(1x)dxdx=1x
s=int_1^3 sqrt(1+(1/x)^2) dxs=∫31√1+(1x)2dx
s=int_1^3 (sqrt(x^2+1))/x dxs=∫31√x2+1xdx
Using Algebraic Substitution
Let z=sqrt(x^2+1)z=√x2+1
and z^2=x^2+1z2=x2+1
Differentiating both sides
2z *dz=2x*dx+02z⋅dz=2x⋅dx+0
2z *dz=2x*dx2z⋅dz=2x⋅dx
dividing both sides by x^2x2
2z *dz=2x*dx2z⋅dz=2x⋅dx
(2z)/x^2 *dz=(2x)/x^2*dx=(2dx)/x 2zx2⋅dz=2xx2⋅dx=2dxx
(2z)/x^2 *dz=(2dx)/x 2zx2⋅dz=2dxx
(2z*dz)/(z^2-1) =(2dx)/x 2z⋅dzz2−1=2dxx
and
(z*dz)/(z^2-1) =(dx)/x z⋅dzz2−1=dxx
Let us go back to
s=int_1^3 (sqrt(x^2+1))/x dx=int_1^3 (sqrt(x^2+1))(dx/x)s=∫31√x2+1xdx=∫31(√x2+1)(dxx)
Replacing now with variable zz
s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=∫31(√x2+1)(dxx)=∫√10√2(z)(z⋅dzz2−1)
Because z=sqrt(x^2+1)" "z=√x2+1
If xx is from x=1x=1 to x=3x=3, then, zz is from z=sqrt2z=√2 to z=sqrt10z=√10
Now we integrate
s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=∫31(√x2+1)(dxx)=∫√10√2(z)(z⋅dzz2−1)
s=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=∫√10√2(z)(z⋅dzz2−1)
s=int_sqrt2^sqrt10 ((z^2*dz)/(z^2-1))s=∫√10√2(z2⋅dzz2−1)
s=int_sqrt2^sqrt10 ((z^2-1+1)*dz)/(z^2-1)s=∫√10√2(z2−1+1)⋅dzz2−1
s=int_sqrt2^sqrt10 ((z^2-1)/(z^2-1)+1/(z^2-1))*dzs=∫√10√2(z2−1z2−1+1z2−1)⋅dz
s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dzs=∫√10√2(1+1z2−1)⋅dz
s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dz=[z+1/2*ln((z-1)/(z+1))]_sqrt2^sqrt10s=∫√10√2(1+1z2−1)⋅dz=[z+12⋅ln(z−1z+1)]√10√2
s=sqrt10+1/2ln((sqrt10-1)/(sqrt10+1))-(sqrt2+1/2ln((sqrt2-1)/(sqrt2+1)))s=√10+12ln(√10−1√10+1)−(√2+12ln(√2−1√2+1))
color(red)(s=2.3019875345832)" "s=2.3019875345832 units
God bless....I hope the explanation is useful.