How do you find the arc length of the curve #y=lnx# over the interval [1,2]?

#y=lnx#

#y'=1/x#

Arc length is given by:

#L=int_1^2sqrt(1+1/x^2)dx#

Rearrange:

#L=int_1^2sqrt(x^2+1)/xdx#

Multiply numerator and denominator by #sqrt(x^2+1)#:

#L=int_1^2(x^2+1)/(xsqrt(x^2+1))dx#

Integration is distributive:

#L=int_1^2x/sqrt(x^2+1)dx+int_1^2 1/(xsqrt(x^2+1))dx#

Apply the substitution #x=tantheta#:

#L=[sqrt(x^2+1)]_ 1^2+intsectheta/tanthetad theta#

Simplify:

#L=sqrt5-sqrt2+intcscthetad theta#

Integrate directly:

#L=sqrt5-sqrt2-[ln|csctheta+cottheta|]#

Rewrite in terms of #tantheta# and #sectheta#:

#L=sqrt5-sqrt2-[ln|(1+sectheta)/tantheta|]#

Reverse the substitution:

#L=sqrt5-sqrt2-[ln((1+sqrt(x^2+1))/x)]_1^2#

Insert the limits of integration:

#L=sqrt5-sqrt2-ln((1+sqrt5)/(2(1+sqrt2)))#

Rearrange for clarity:

#L=sqrt5-sqrt2+ln2-ln((1+sqrt5)/(1+sqrt2))#

The arc length of a function #y=f(x)# over the interval #[a,b]# is given by #L=int_a^bsqrt(1+(dy/dx)^2)dx#.
#y=lnx->dy/dx=1/x#
#:.L=int_1^2sqrt(1+(1/x)^2)dx#
#L=int_1^2sqrt(1+1/x^2)dx#
I will use this solution #intsqrt(1+1/x^2)dx=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|+C# courtesy of mason m (https://socratic.org/questions/how-do-you-evaluate-the-integral-int-sqrt-1-1-x-2) to find that
#L=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|]_1^2#
#:.L=sqrt5+1/2ln((sqrt5-1)/(sqrt5+1))-sqrt2-1/2ln((sqrt2-1)/(sqrt2+1))~~1.222#