Question

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

Answer 1

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

Explanation:

`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`

Answer 2

`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`