Question

Evaluate the integrals in Exercises 31–78.75. ∫(from -2 to -1)2dv/(v²+4v+5)

Accepted Answer

Start by examining the integral $$ \int_{-2}^{-1} \frac{2}{v^{2} + 4v + 5} \, dv . $$Notice that the denominator is a quadratic expression, so the first step is to complete the square for the quadratic $$ v^{2} + 4v + 5 . $$Rewrite the quadratic by completing the square: $$ v^{2} + 4v + 5 = (v^{2} + 4v + 4) + 1 = (v + 2)^{2} + 1 . $$This simplifies the integral to $$ \int_{-2}^{-1} \frac{2}{(v + 2)^{2} + 1} \, dv . $$Make a substitution to simplify the integral further. Let $$ u = v + 2 $$, which implies $$ du = dv . $$Change the limits of integration accordingly: when $$ v = -2 $$, $$ u = 0 $$; when $$ v = -1 $$, $$ u = 1 . $$The integral becomes $$ \int_{0}^{1} \frac{2}{u^{2} + 1} \, du . $$Recognize that the integral $$ \int \frac{1}{u^{2} + 1} \, du $$ corresponds to the inverse tangent function $$ \arctan(u) . $$Therefore, the integral can be expressed as $$ 2 \int_{0}^{1} \frac{1}{u^{2} + 1} \, du = 2 [ \arctan(u) ]_{0}^{1} . $$Evaluate the definite integral by substituting the limits into the inverse tangent function: $$ 2 ( \arctan(1) - \arctan(0) ) . $$This expression represents the value of the original integral.

Evaluate the integrals in Exercises 31–78.
75. ∫(from -2 to -1)2dv/(v²+4v+5)

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Key Concepts

Definite Integrals

Completing the Square

Integration of Rational Functions