Integration with sqrt in denominator

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SUMMARY

The integral of the function with a square-rooted denominator, specifically ∫(1/√(3x-x²)) dx, can be solved using trigonometric substitution. The expression 3x-x² can be rewritten as (3/2)² - (x - (3/2))², which facilitates the substitution. Two effective substitutions are suggested: letting (x - (3/2)) = (3/2)sin(t) or (x - (3/2)) = (3/2)sinh(t). These methods simplify the integration process significantly.

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I wonder how would I get out the integral when the denominator is square-rooted.

∫[itex]\frac{1}{\sqrt{3x-x^2}}[/itex] dx
 
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3x-x² = (3/2)²-(x -(3/2) )²
Let (x -(3/2) )= (3/2)*sin(t)
or let (x -(3/2) )= (3/2)*sinh(t)
 

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