I need help with a calculus 2 exercise, with a good explanation to what I'm trying to figure out.The problem involves finding the arc length when y = (x^2)/2 - (lnx/4) given 2≤ x ≤ 4So I know the formula involves L = Integral from a to b of sqrt(1+(f'(x))^2)dx And I took the the derivative of y = (x^2)/2 - (lnx/4)And got = x - (1/4x) I then kept reducing it to suit the formula 1 + ( x - 1/4x)^2I added the alike terms1 + x^2 - 1/2 + 1/16x^2x^2 + 1/2 + 1/16x^2AND then comes my question!So it is supposed to become this afterwards: (x + 1/4x) ^ 2But how did that happen, I don't understand how I can reduce it to a square, please help me figure it out.

Explanation:When writing rational expressions, you need to be aware that ...   1/4x = (1/4)x ≠ 1/(4x)Parentheses around the denominator are required, unless you're typesetting the expression and can use a fraction bar for grouping.The derivative of the curve expression is ...   y' = x - 1/(4x) . . . . . parentheses added to what you wroteand the expression (1 -(y')^2) can be written ...   1 -(y')^2 = x^2 +1/2 +1/(16x^2) . . . . . parentheses added to what you wroteThe first and last terms of this trinomial are both perfect squares, so you might suspect the whole trinomial is a perfect square. You recall that ...   (a +b)^2 = a^2 + 2ab + b^2This is a good "pattern" to remember. Using it is a matter of pattern recognition, as is the case with a lot of math.Here, you have ...   a = x   b = 1/(4x)In order for your trinomial to be a perfect square, the product 2ab must equal the middle term of your trinomial. (Spoiler: it does.)   2ab = 2(x)(1/(4x)) = (2x)/(4x) = 1/2 . . . . . matches the middle term of 1 -(y')^2Hence your trinomial can be written as the square ...   1 -(y')^2 = (x +1/(4x))^2_____This is convenient because you want to integrate the square root of this. Your integral then becomes ...[tex]\displaystyle\int\limits_{2}^{4}{\left(x+\frac{1}{4x}\right)\,dx[/tex]