Basically, you use the arc length formula:

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

And you have to simplify down to a perfect square and then take the square root. The simplification is the hard part. Afterwards it's very simple (keep reading).

*You can find the derivation for the arc length at the bottom if you don't remember it or don't have it derived.*

#f(x) = (x^2/4) - 1/2lnx#

#s = int_1^e sqrt(1 + (x/2 - 1/(2x))^2)dx#

Multiply stuff out and add everything together:

#= int_1^e sqrt(x^2/4 + 1/2 + 1/(4x^2))dx#

Small trick with multiplication:

#= 1/2int_1^e 2sqrt(x^2/4 + 1/2 + 1/(4x^2))dx#

Another small trick with square roots:

#= 1/2int_1^e cancel(sqrt4)sqrt(x^2/cancel(4) + cancel(1/2)^(2) + 1/(cancel(4)x^2))dx#

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

Making it look nicer:

#= 1/2int_1^e sqrt((x^4 + 2x^2 + 1)/(x^2))dx#

Since we are in a positive domain, all this works out to be:

#int_1^e sqrt(1 + (x/2 - 1/(2x))^2)dx = 1/2int_1^e sqrt(x^4 + 2x^2 + 1)/xdx#

Alright, this looks good to go. Thankfully, we have a perfect square now, AND we know that #x > 0#, always, in this domain.

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

#= 1/2int_1^e x + 1/xdx#

#= 1/2 [x^2/2 + ln|x|]|_(1)^(e)#

#= 1/2[((e)^2/2 + cancel(ln|e|)^1) - ((1)^2/2 + cancel(ln|1|)^0)]#

#= 1/2(e^2/2 + 1/2)#

#= 1/4(e^2 + 1) ~~ 2.09726#

**Derivation**

The arc length can be derived, if you forget the (rather simple) formula. Since it really is just *an accumulation of really short straight lines*, you can use the distance formula and modify it in such a way that it works for a specific variable using the definition of the derivative.

#D = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2)#

#= sqrt((Deltax)^2 + (Deltay)^2)#

We can let #x# vary as it may, because we're assuming that the arc length is of a function that more than likely acts according to the "**vertical line test**", which means a **continuously increasing** #x# (never decreasing AND increasing in the same graph) but a **varying** #y#. What we need to do is modify how #Deltay# is written to allow for any such variation in #y# that matches how our function behaves. Recall the Leibnitz notation of the derivative:

#(dy)/(dx) = (Deltay)/(Deltax)#, but varying over infinitely small intervals

Thus:

#(Deltay)^2 = ((Deltay)/(Deltax))^2*(Deltax)^2#

#D = sqrt((Deltax)^2 + ((Deltay)/(Deltax))^2*(Deltax)^2)#

Since #(Deltax)^2# is *always* increasing, it will *always* be positive on a graph where #x# increases from left to right, so there's no need to put it as #|Deltax|# when pulling it out of the square root. Now let's call the arc length #s#, like in #s = rtheta#, and add a #sum# symbol to accumulate the tiny, short straight lines into an arc length:

#s = sumsqrt(1 + ((Deltay)/(Deltax))^2)*Deltax#

Finally, this can be converted to an integral by using #Deltax# as #dx#, etc., and replacing the sum with an integral symbol:

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