Hint: Consider the equilateral triangle with base=10v, and height=3400. So, half base=5v.
Check that tan(15)=5v/3400, v is in mt/sec, angle in degrees.
Substitute, x= 1/z. Then, apart form the constants, the integral becomes of the form 1/sqrt(4/9-z^2).
The final answer should be (1/3).arcCos{3/(2x)} +C [or, -(1/3).arcSin{3/(2x)} +C].
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