Remember that arc length is given by #int_a^bsqrt(1+(dy/dx)^2)dx#.

We can see that we need the derivative of the function in question, so let's get to that first:

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

#f'(x)=2x#

Next step is to plug this into the formula:

#int_0^1sqrt(1+(2x)^2)dx#

Before we do anything, we should note that this integral merits a trig substitution. It resembles a right triangle (see below).

From the picture, we know that #tan(theta)=(2x)/1=2x#. That means #tan(theta)/2=x#, and furthermore, #(dx)/(d theta)=sec^2(theta)/2#. Multiplying both sides by #d theta# tells us #dx=sec^2(theta)/2d theta#.

Making these substitutions into the integral,

#int_0^1sqrt(1+(2(tan(theta)/2))^2)sec^2(theta)/2d theta#

Note that while the stuff inside the integral is in terms of #theta# now, the limits of integration (0 and 1) are still in terms of #x#. To fix this issue, recall that #tan(theta)=2x#. Taking the inverse tangent of both sides yields #theta=tan^(-1)(2x)#. Now we can get #x=0# and #x=1# in terms of #theta#:

#theta=tan^(-1)(2(0))->theta=tan^(-1)(0)=0#

#theta=tan^(-1)(2(1))->theta=tan^(-1)(2)~~1.107#

Unfortunately, there is no exact answer for #tan^(-1)(2)#, so we'll have to live with an approximation. Our integral finally becomes

#int_0^1.107sqrt(1+(2(tan(theta)/2))^2)sec^2(theta)/2d theta#

Simplifying:

#1/2int_0^1.107sqrt(1+(tan(theta))^2)sec^2(theta)d theta#

#1/2int_0^1.107sqrt(1+tan^2(theta))sec^2(theta)d theta#

If you remember your trig well, you'll know that #1+tan^2(theta)=sec^2(theta)#.

#1/2int_0^1.107sqrt(sec^2(theta))sec^2(theta)d theta#

#1/2int_0^1.107sec^3(theta)d theta#

Evaluating this integral is unnecessary because we can consult a table of integrals to get a result.

After doing this, we see that #intsec^3(theta)d theta=1/2sec(theta)tan(theta)+1/2lnabs(sec(theta)+tan(theta))#. The final step is finding what this is on the interval #[0, 1.107]#:

#1/2[1/2sec(theta)tan(theta)+1/2lnabs(sec(theta)+tan(theta))]_0^1.107#

#=1/2(1/2sec(1.107)tan(1.107)+1/2lnabs(sec(1.107)+tan(1.107)))-1/2(1/2sec(0)tan(0)+1/2lnabs(sec(0)+tan(0)))#

#=1.478-0=1.478#