How to integrate a function within a square root

seanm924
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Homework Statement


Find the length of x=y^3/30+5/(2y) on 3 <= y <= 5


Homework Equations



L = integral sqrt(1+(dx/dy)^2)dx

The Attempt at a Solution



I got to the integral upper bound = 5, lower bound = 3, sqrt(y^4/100+25/(4y^4)+1/2)dy. How do I actually solve the integral? It would be find if there wasn't a square root.
 
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Factor out \frac{1}{100y^4} inside the root and see if you can go from there.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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