How do you evaluate the integral #int sqrt(1+1/x^2)#?

#intsqrt(1+1/x^2)dx=intsqrt((x^2+1)/x^2)dx=intsqrt(x^2+1)/xdx#

Letting #u=sqrt(x^2+1)# reveals that #du=x/sqrt(x^2+1)dx#. Then the integral can be manipulated to become:

#intsqrt(x^2+1)/xdx=int(x(x^2+1))/(x^2sqrt(x^2+1))dx=int(x^2+1)/x^2(x/sqrt(x^2+1)dx)#

Note that #u^2=x^2+1# and #u^2-1=x^2#:

#int(x^2+1)/x^2(x/sqrt(x^2+1)dx)=intu^2/(u^2-1)du#

Rewriting the integrand as #(u^2-1+1)/(u^2-1)=1+1/(u^2-1)#:

#intu^2/(u^2-1)du=intdu+int1/(u^2-1)du=u+int1/(u^2-1)du#

You can perform partial fraction decomposition on #1/(u^2-1)# to see that #1/(u^2-1)=1/(2(u-1))-1/(2(u+1))#:

#u+int1/(u^2-1)du=u+1/2int1/(u-1)du-1/2int1/(u+1)du#

Both of which you could perform a substitution on, or just recognize them for natural log integrals:

#u+1/2int1/(u-1)du-1/2int1/(u+1)du=u+1/2lnabs(u-1)-1/2lnabs(u+1)#

Combining:

#=u+1/2lnabs(u-1)-1/2lnabs(u+1)=u+1/2lnabs((u-1)/(u+1))#

Since #u=sqrt(x^2+1)#:

#u+1/2lnabs((u-1)/(u+1))=sqrt(x^2+1)+1/2lnabs((sqrt(x^2+1)-1)/(sqrt(x^2+1)+1))#

Or:

#intsqrt(1+1/x^2)dx=sqrt(x^2+1)+1/2lnabs((sqrt(x^2+1)-1)/(sqrt(x^2+1)+1))+C#

#I=intsqrt(1+1/x^2)dx#

Let #x=cottheta#. This implies that #dx=-csc^2thetad theta# and that #1+1/x^2=1+tan^2theta=sec^2theta#. Then:

#I=intsqrt(sec^2theta)(-csc^2thetad theta)=-intsectheta(1+cot^2theta)d theta#

#color(white)I=-int(sectheta+cotthetacsctheta)d theta=-lnabs(sectheta+tantheta)+csctheta#

Rewriting in terms of #cottheta#:

#I=-lnabs(sqrt(1+1/cot^2theta)+1/cottheta)+sqrt(1+cot^2theta)#

#color(white)I=-lnabs(sqrt(1+1/x^2)+1/x)+sqrt(1+x^2)#

#color(white)I=lnabs(x/(1+sqrt(1+x^2)))+sqrt(1+x^2)+C#