Repeated linear factors Evaluate the following integrals.

Problem 25EAnswer;Step-1Proper fraction definition ; In a rational fraction , if the degree of f(x) < the degree of g(x) , then the rational fraction is called a proper fraction. The sum of two proper fractions is a proper fraction. Example; Partial fractions Depending upon the nature of factors of Denominator ;1. When the denominator has non-repeated linear factors; A non - repeated linear factor (x-a) of denominator corresponds a partial fraction of the form , where A is a constant to be determined’ If g(x) = (x-a)(x-b)(x-c)............(x-n), then we assume that = ++ +...............+Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a,b ,c ….n.Step-21. When the denominator has repeated linear factors A repeated linear factor of denominator corresponds partial fractions of the form ; = + ++......................+Where A, B, C,............N are constants which can be determined by equating the numerator of L.H.Sto the numerator of R.H.S , and substituting x = a, we get N. Example; Step-3 The given integral is ; dx………………(1) Here the given integrand is of the form , and f(x) < g(x). Therefore , the given fraction is a proper fraction , and the denominator has repeated linear factors. Thus , from the above step the given fraction can be written as; = ++ = …………..(2) Equating the numerator of L.H.S to the numerator of R.H.S Then , x -5 = A+Ax +Bx+B +C = (A+C) +x(A+B) +B Therefore, A+C = 0 , A+B =1 and B = -5 , since equating the coefficients. Solving , A+C = 0 , A+B =1 and B = -5 then A= 1-B = 1-(-5) = 1+5 = 6 A+C =0 , then C = -A = -6 Thus , A = 6 , B = -5 and C =-6…………….(3) From(2) , (3) = ++ = ++ = --…………..(4) Therefore, from(1) and (4) dx = = dx -dx = 6dx -dx = 6ln(|x|) - - 6ln(|x+1|)+ C , since = +C , dx = ln(x)+c = 6 ln(||) - +c , ln(a) - ln(b) = ln(a/b) Therefore , dx = 6 ln(||) - +c