How do I find the length of this curve? I have a curve $ y = \frac{3}{4}x^\frac{4}{3} - \frac{3}{8}x^\frac{2}{3} + 7$. I’m required to find out the length of this curve from $x=1$ to $x=8$. 
 Using the formula of length of curve,I am trying to evaluate the following integral —
$\int_1^8 \sqrt{1 + (x^\frac{1}{3} - \frac{1}{4x^\frac{1}{3}})^2} dx$ but I’m not sure how to proceed further. Please help. 
 A: Hint: We note $x^{1/3}$ appearing twice in the integral. While not a sure-fire solution in general, here this works out partly: make the $u$-substitution $u=x^{1/3}$. Following this, what's under the radical will be a perfect square trinomial, and the result becomes easier to obtain.

Solution:
We wish to calculate
$$\mathcal I = \int_1^8 \sqrt{1 + \left(x^{1/3} - \frac{1}{4x^{1/3}} \right)^2} dx$$
Let $u=x^{1/3}$. Then $du = \frac 1 3 x^{-2/3}dx = \frac 1 3 u^{-2}dx$, giving us $dx=3u^2$. We change our bounds by noting that $u(1) = 1, u(8) = 2$. Thus,
$$\mathcal I = 3 \int_1^2 \sqrt{1 + \left(u - \frac{1}{4u} \right)^2} \cdot u^2du$$
Expand the square term in the radical:
$$\mathcal I = 3\int_1^2 \sqrt{u^2 - 2u\frac{1}{4u} + \frac{1}{16u^2}+1} \cdot u^2du$$
Simplify, and bring the $u^2$ term into the radical by noting $u^2 = \sqrt{u^4}$:
$$\mathcal I =3 \int_1^2 \sqrt{u^6 + \frac 1 2 u^4 + \frac{1}{16} u^2} \; du$$
Now, if we hope there is some stroke of luck, the radical will be a perfect square trinomial, so we test this theory. To see if a trinomial is a perfect square, note they take the form $a^2 + 2ab + b^2 = (a+b)^2$, so we work with the square roots of the first term ($u^3$) and the last ($u/4$) and test:
$$\left( u^3 + \frac u 4 \right)^2 = (u^3)^2 + 2u^3 \frac u 4 + \left( \frac u 4 \right)^2 = u^6 + \frac{u^4} 2 + \frac{u^2}{16}$$
Success!
$$\mathcal I = 3\int_1^2 u^3 + \frac u 4 \; du$$
Being a polynomial, evaluating it from this point will be easy for you.
