Question

Evaluate the integrals in Exercises 77–90.77. ∫dx/√(-x²+4x-3)

Accepted Answer

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$.

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

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Key Concepts

Completing the Square

Integration of Functions Involving Square Roots

Trigonometric Substitution