Calculus
posted by Michael .
Evaluate the integral.
The integral from the square root of three over three to the square root of three of the function 6/(t^2+1)

This is one of your standard integrals:
∫dx/(x^2+1) = arctan(x)
now, what angle θ has tanθ = √3/3?
θ = pi/6
Now go and crank on it. If you get stuck, come on back.
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