Evaluate the integral: I = integral dx/(sqrt(-2x²+5x-2))

Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{-2x^2+5x-2}}. $$

Explanation of Method

Completing the square converts the radical into the standard $\sqrt{a^2-u^2}$ form, resolved directly by the arcsine formula.

$$ -2x^2+5x-2 = -2\left(x^2-\frac52x+1\right) = -2\left[\left(x-\frac54\right)^2-\frac{9}{16}\right] = \frac98 - 2\left(x-\frac54\right)^2. $$

So:

$$ \sqrt{-2x^2+5x-2} = \sqrt2\,\sqrt{\frac{9}{16}-\left(x-\frac54\right)^2}. $$

Let $u=x-\tfrac54$, $du=dx$, $a=\tfrac34$:

$$ I = \frac{1}{\sqrt2}\int \frac{du}{\sqrt{a^2-u^2}}. $$

$$ I = \frac{1}{\sqrt2}\arcsin\frac ua + C = \frac{1}{\sqrt2}\arcsin\left(\frac{4u}{3}\right) + C. $$

With $u=x-\tfrac54$, so $4u=4x-5$:

$$ I = \frac{1}{\sqrt2}\arcsin\left(\frac{4x-5}{3}\right) + C = \frac{\sqrt2}{2}\arcsin\left(\frac{4x-5}{3}\right) + C. $$

Correction: the source introduced an unmotivated variable $z=\sqrt{9/8-(x-5/4)^2}$ and a "rational substitution" step that was never carried through to a valid result — the stated final answer, $-\sqrt2\arctan\sqrt{\tfrac{-2x^2+5x-2}{2(x-2)}}$, does not differentiate back to the integrand and was discarded. Since $-2x^2+5x-2$ is a downward-opening quadratic with real roots ($x=\tfrac12,2$), completing the square and applying the standard $\int du/\sqrt{a^2-u^2}=\arcsin(u/a)$ formula is the direct, reliable route.
The domain requires $-2x^2+5x-2\geq0$, i.e. $\tfrac12\leq x\leq2$ (the roots of $-2x^2+5x-2=-(2x-1)(x-2)$).
The result was verified by differentiating $\tfrac{1}{\sqrt2}\arcsin\!\left(\tfrac{4x-5}{3}\right)$: this gives $\tfrac{1}{\sqrt2}\cdot\tfrac{4/3}{\sqrt{1-(4x-5)^2/9}}=\tfrac{4}{3\sqrt2}\cdot\tfrac{3}{\sqrt{9-(4x-5)^2}}=\tfrac{4}{\sqrt2\sqrt{9-(4x-5)^2}}$; since $9-(4x-5)^2=8(-2x^2+5x-2)$, this simplifies to $\tfrac{4}{\sqrt2\cdot2\sqrt2\sqrt{-2x^2+5x-2}}=\tfrac{1}{\sqrt{-2x^2+5x-2}}$, matching the integrand.

The solution to the integral is:

$$ I = \frac{\sqrt2}{2}\arcsin\left(\frac{4x-5}{3}\right) + C. $$