Calc1231W5T2 - Integrals of rational functions

Friday, 24 August 2018

1:05 PM

Untitled picture.png Machine generated alternative text: The integral of a function of the form Hence 5 dc = In(aös(5x+2)) Note: the Maple syntax for In(lxl) is In(abs (x) ) 1 dc Ink Drawings Ink Drawings Ink Drawings Ink Drawings                   
  
Untitled picture.png Machine generated alternative text: The integral 1 1 dc. dc , as in the previous questions. But with a bit of algebraic manipulation we can make it into two integrals of this form. This is because is not in the form f@) 1 for some constants A and B Let's find the value of these constants. First multiply both sides of this equation by @ — 4)(x — 3) . This gives • Setting x • Setting x 4 in this equation means that A 3 in this equation means that B Now the functions are in the required form x—4' x—3 This makes the original integral I easy to calculate as the sum of f@) and 1 dc 1 dc In(abs(x-4))  
Ink Drawings Ink Drawings Untitled picture.png Machine generated alternative text: What about an integral of the form 1 1 The plan is the same, but the execution is slightly different. We split up the integrand into three functions that we can integrate individually: 1 @ — l)(x — where A, B and C are constants. Ifwe multiply this equation by @ — 1)@ — then — + B(x — 1)@ — 4) + ¯ 1) Now lets find the values of these constants: • Setting x = 1 gives A = 1/9 • Setting x = 4 gives C = 1/3 We have to work a little harder to find B. Comparing coefficients of Hence our original integral I is the sum of the three integrals: 1 1 dc 1 dc is probably the easiest way of finding B, since we only have to solve Ink Drawings Ink Drawings                

Untitled picture.png Machine generated alternative text: Using the techniques from this topic, it is possible to evaluate integrals such as dc = J +25 and dc J x2 — 25 0åt+C Untitled picture.png Machine generated alternative text: It's not always possible to factorize a quadratic into real linear terms. For example does not factorize into real linear terms since the discriminant is less than zero. In this case we complete the square to write p@) as a sum of two squares: where Hence we can integrate by first completing the square Then using the standard integral formula we calculate Note: the discriminant of ax2 + bx c is A = b2 4 + -4m + 20 1 dc 1 1 1 .T2 + 4X + 20 1 du 2 4ac 1 1 and the Maple syntax for tan¯ 1 I(au) is arctan (x) Ink Drawings Ink Drawings Ink Drawings Ink Drawings                 
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