Evaluate the integral: I = integral dx/(sqrt(4+2x-x²))

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{4+2x-x^2}}. $$

Step 1 · Complete the square

$$ 4+2x-x^2 = -\left(x^2-2x-4\right) = -\left[(x-1)^2-1-4\right] = 5-(x-1)^2. $$

Step 2 · Substitute

Let $u=x-1$, $du=dx$:

$$ I = \int \frac{du}{\sqrt{5-u^2}}. $$

Step 3 · Apply the standard arcsine formula

With $a=\sqrt5$:

$$ I = \arcsin\frac{u}{\sqrt5} + C. $$

Step 4 · Substitute back

With $u=x-1$:

$$ I = \arcsin\left(\frac{x-1}{\sqrt5}\right) + C. $$

Summary Table

Additional Notes

• The domain requires $1-\sqrt5\leq x\leq1+\sqrt5$, where $4+2x-x^2\geq0$.
• This closes out the run of $\sqrt{a^2-u^2}$-type radicals (Problems 181, 194, 196, 200) that all resolve to arcsine, contrasting with the $\sqrt{u^2\pm a^2}$-type radicals (Problems 182, 195, 197–199) that resolve to logarithms.
• The result can be checked by differentiating $\arcsin\!\left(\tfrac{x-1}{\sqrt5}\right)$: this gives $\dfrac{1/\sqrt5}{\sqrt{1-(x-1)^2/5}}=\dfrac{1}{\sqrt{5-(x-1)^2}}=\dfrac{1}{\sqrt{4+2x-x^2}}$, matching the integrand.