Evaluate the integrals in Exercises 31–78.
73. ∫dx/√(-2x-x²)

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Rewrite the expression inside the square root to a more recognizable form by completing the square for the quadratic expression: \(-2x - x^2\).

Start by factoring out the negative sign: \(- (x^2 + 2x)\), then complete the square inside the parentheses: \(x^2 + 2x = (x + 1)^2 - 1\).

Substitute back to get the integrand as \(\frac{1}{\sqrt{- ( (x + 1)^2 - 1 )}} = \frac{1}{\sqrt{1 - (x + 1)^2}}\).

Recognize that the integral now has the form \(\int \frac{dx}{\sqrt{a^2 - u^2}}\), which is a standard integral with solution involving inverse trigonometric functions.

Use the substitution \(u = x + 1\) and apply the formula \(\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C\) to express the integral in terms of \(x\).

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