Evaluate the integrals in Exercises 77–90.
77. ∫dx/√(-x²+4x-3)

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Start by rewriting the expression under the square root to a more recognizable form. The integrand is $\frac{1}{\sqrt{-x^{2} + 4x - 3}}$. Notice the quadratic inside the square root: $-x^{2} + 4x - 3$.

Complete the square for the quadratic expression inside the square root. Factor out the negative sign first: $-(x^{2} - 4x + 3)$. Then complete the square for $x^{2} - 4x + 3$ by finding a perfect square trinomial.

Recall that $x^{2} - 4x + 3 = (x - 2)^{2} - 1$ because $(x - 2)^{2} = x^{2} - 4x + 4$. Substitute this back to get $-( (x - 2)^{2} - 1 ) = 1 - (x - 2)^{2}$ inside the square root.

Rewrite the integral as $\int \frac{dx}{\sqrt{1 - (x - 2)^{2}}}$. This is a standard form that resembles the integral of $\frac{1}{\sqrt{a^{2} - u^{2}}}$, which is related to the inverse sine function.

Use the substitution $u = x - 2$, so $du = dx$. The integral becomes $\int \frac{du}{\sqrt{1 - u^{2}}}$. Recognize this as the integral of $\frac{1}{\sqrt{1 - u^{2}}}$, whose antiderivative is $\arcsin(u) + C$. Finally, substitute back $u = x - 2$ to express the answer in terms of $x$.

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