![\\ I = \int_{5}^{\infty} \frac{6}{x^2 - x - 2}dx \\ \\ I = \int_{5}^{\infty} \frac{6}{x^2 - 2*(1/2)*x + (1/4) - 1/4 - 2}dx \\ \\ I = \int_{5}^{\infty} \frac{6}{(x- 1/2)^2 - 9/4}dx \\ \\ Substitute: \ \ x - 1/2 = u \\ \\ dx = du \\ \\ I = \int_{5}^{\infty} \frac{6}{u^2 - 9/4}du \\ \\ I = \int_{5}^{\infty} \frac{24}{4u^2 - 9}du \\ \\ I = 24 \int_{5}^{\infty} \frac{1}{(2u - 3)(2u + 3)}du \\ \\ \frac{1}{(2u - 3)(2u + 3)} = \frac{A}{2u - 3} + \frac{B}{2u + 3} \\ \\ 1 = A(2u + 3) + B(2u - 3) \\ \\ When \ \ u = 3/2, \ \ then \ \ A = \frac{1}{6} \\ \\ When \ \ u = -3/2, \ \ then \ \ B = \frac{-1}{6} \\ \\ So, \\ \\ I = 24 \int_{5}^{\infty} [\frac{1}{6(2u - 3)} - \frac{1}{6(2u + 3)}]du \\ \\ I = 4 \int_{5}^{\infty} [\frac{1}{(2u - 3)} - \frac{1}{(2u + 3)}]du \\](http://img.homeworklib.com/questions/b76dc220-e68e-11ea-b441-a133ccdbaa85.png?x-oss-process=image/resize,w_560)
![\\ I = 4 \int_{5}^{\infty} [\frac{1}{(2u - 3)} - \frac{1}{(2u + 3)}]du \\ \\ I = 4[\frac{ln (2u - 3)}{2} - \frac{ln (2u + 3)}{2}]_{5}^{\infty} \\ \\ I = 4[\frac{ln (2(x - 1/2) - 3)}{2} - \frac{ln (2*(x - 1/2) + 3)}{2}]_{5}^{\infty} \\ \\ I = 4[\frac{ln (2x - 4)}{2} - \frac{ln (2x + 2)}{2}]_{5}^{\infty} \\ \\ I = 2[ln (\frac{(2x - 4)}{(2x + 2)})]_{5}^{\infty} \\ \\ I = 2[0 - ln (\frac{(2*5 - 4)}{(2*5 + 2)})] \\ \\ \lim_{x \rightarrow \infty} ln (\frac{(2x - 4)}{(2x + 2)}) =ln [ \lim_{x \rightarrow \infty} \frac{(2x - 4)}{(2x + 2)}] = ln (1) = 0 \\ \\ I = 2[0 - ln (\frac{1}{2})] \\ \\ I = {\color{Red} 2*ln 2} \\](http://img.homeworklib.com/questions/b7c2d100-e68e-11ea-8c65-2733ae3c940c.png?x-oss-process=image/resize,w_560)
Evaluate the following integrals
Evaluate the following integrals: 3 1 32/2 7 (i) dx 6°2712 dx x²x² + 4 (4x² +9312
1 (a) Evaluate the integral sz? dx. (32-7) 5 2 7 (b)if s f(x)dx = 3 and 5 f(x)dt = 4, find S f(x)dr. 5
please help
Evaluate the integral. 2 (2x + 2)dx 3 2 Sizx+2)dx= (2x + 2)dx = (Simplify your answer.) 3
evaluate
2 (a) (x2 + 1)2 dx
Evaluate ∫C(2x - y) dx + (x + 3y)dy C: arc on y=x5/2 from (0, 0) to (4, 32) _______
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Evaluate each integral (a) (x² + x) dx (b) 6.** (secx + tanx)2 dx
(Calc 2)
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Evaluate the definite integral by the limit definition. + 2) dx
8.)Evaluate: a.) S 4x® dx. b.) ſ 4x2Vx3 + 3 dx. c.) S(x3 +1 )2 dx.