How do you find the length of the curve for #y= ln(1-x²)# for (0, 1/2)?

The formula for the arc length of the curve #y# on the interval #[a,b]# is given by:

#s=int_a^bsqrt(1+(dy/dx)^2)dx#

Here, where #y=ln(1-x^2)#, then #dy/dx=(-2x)/(1-x^2)=(2x)/(x^2-1)#.

Thus, the arc length in question is:

#s=int_0^(1//2)sqrt(1+((2x)/(x^2-1))^2)dx#

#s=int_0^(1//2)sqrt(((x^2-1)^2+(2x)^2)/(x^2-1)^2)dx#

#s=int_0^(1//2)sqrt(x^4-2x^2+1+4x^2)/(x^2-1)dx#

#s=int_0^(1//2)sqrt(x^4+2x^2+1)/(x^2-1)dx#

Note this is factorable:

#s=int_0^(1//2)sqrt((x^2+1)^2)/(x^2-1)dx#

#s=int_0^(1//2)(x^2+1)/(x^2-1)dx#

Rewriting:

#s=int_0^(1//2)(x^2-1+2)/(x^2-1)dx#

#s=int_0^(1//2)(1+2/(x^2-1))dx#

Perform partial fractions on the second piece:

#2/(x^2-1)=A/(x-1)+B/(x+1)#

#2=A(x+1)+B(x-1)#

Letting #x=1# reveals that

#2=2A" "=>" "A=1#

And letting #x=-1# shows that

#2=-2B" "=>" "B=-1#

So

#2/(x^2-1)=1/(x-1)-1/(x+1)#

Then

#s=int_0^(1//2)(1+1/(x-1)-1/(x+1))dx#

Which are all easily integrateable:

#s=[x+lnabs(x-1)-lnabs(x+1)]_0^(1//2)#

#s=[x+lnabs((x-1)/(x+1))]_0^(1//2)#

#s=(1/2+lnabs((1/2-1)/(1/2+1)))-(0+lnabs((0-1)/(0+1)))#

#s=1/2+lnabs((-1/2)/(3/2))+lnabs(-1)#

#s=1/2+ln(1/3)#

Or

#s=1/2-ln3#