Evaluate the integrals in Exercises 77–90.
84. ∫(from 2 to 4)2dx/(x²-6x+10)

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First, recognize that the integral is of the form $\int_{2}^{4} \frac{2}{x^{2} - 6x + 10} \, dx$. The denominator is a quadratic expression, so start by completing the square for the quadratic $x^{2} - 6x + 10$.

Rewrite the quadratic as $x^{2} - 6x + 10 = (x^{2} - 6x + 9) + 1 = (x - 3)^{2} + 1$. This simplifies the integral to $\int_{2}^{4} \frac{2}{(x - 3)^{2} + 1} \, dx$.

Make a substitution to simplify the integral: let $u = x - 3$, so that $du = dx$. Change the limits of integration accordingly: when $x = 2$, $u = 2 - 3 = -1$; when $x = 4$, $u = 4 - 3 = 1$.

Rewrite the integral in terms of $u$: $\int_{-1}^{1} \frac{2}{u^{2} + 1} \, du$. Recognize that the integral of $\frac{1}{u^{2} + 1}$ is $\arctan(u)$.

Use the integral formula $\int \frac{1}{u^{2} + 1} \, du = \arctan(u) + C$ to write the integral as $2 \int_{-1}^{1} \frac{1}{u^{2} + 1} \, du = 2 [\arctan(u)]_{-1}^{1}$. Evaluate this expression by substituting the limits.

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A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating a definite integral involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.

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