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**Problem statement:**

Find the arc length of the curve defined by x = √t and y = 3t -1 on the interval 0 < t < 1

**attempted solution:**

dx/dt = 1/2t

^{-1/2}, dy/dt = 3 and dx = dt/ 2√t

dy/dx = 6√t

length = ∫

_{0}

^{1}√(1 + (6√t)

^{2}) .dt/ 2√t

= ∫

_{0}

^{1}√1 + 36t) dt/2√t

now I'm stuck with a product that is very difficult to solve using integration by parts or by trig substitution. Please shed some light.

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