evaluate the integral x+21/x^2+20x+104 Solution ? 1 / [ x * (x + 1) * (x + 2) ]
ID: 3140058 • Letter: E
Question
evaluate the integral x+21/x^2+20x+104Explanation / Answer
? 1 / [ x * (x + 1) * (x + 2) ] ---> partial fraction 1 / [ x * (x + 1) * (x + 2) ] = [ A / x ] + [ B / (x + 1) ] + [ C / (x + 2) ] --> times x * (x + 1) * (x + 2) 1 = A(x + 1) * (x + 2) + B (x + 2) * x + C (x + 1) * x 1 = A(x^2 + 3x + 2) + B (x^2 + 2x) + C (x^2 + x) 1 = x^2A + 3xA + 2A + x^2B + 2xB + x^2C + xC 1 = 2A ----> A = 1/2 [ 0 = 3A + 2B + C ] x . . -3*(1/2) - 2B = C. . . .(-3/2) - 2B = C [ 0 = A + B + C ] x^2 . . . -(1/2) - B = C C = C -(1/2) - B = (-3/2) - 2B -(1/2) + (3/2) = B - 2B (2/2) = - B -1 = B . . .. . C = -1/2 - -1 -----> 1/2 = C 1 / [ x * (x + 1) * (x + 2) ] = [ (1/2) / x ] + [ -1 / (x + 1) ] + [ (1/2) / (x + 2) ] --> integrate ? 1 / [ x * (x + 1) * (x + 2) ] = (1/2) * ln(x) - ln(x + 1) + (1/2) * ln(x + 2) (1/2) * ln(x) - ln(x + 1) + (1/2) * ln(x + 2) ----> from 1 to 8 (1/2) * ( ln(8) - ln(1) ) - ( ln(8 + 1) - ln(1 + 1) ) + (1/2) * ( ln(8 + 2) - ln(1 + 2) ) (1/2) * ( ln(8) - 0 ) - ( ln(8) - ln(2) ) + (1/2) * ( ln(8) - ln(3) ) lim ln(x) = pplying L'Hopital's Rule x--->8 ln(x) ---> apply 1/x lim 1/x = 1/8 = 0 x--->8 lim ln(x + 1) = pplying L'Hopital's Rule x--->8 ln(x + 1) ---> apply 1/(x + 1) lim 1/(x + 1) = 1/(8 + 1) = 1/8 = 0 x--->8 lim ln(x + 2) = pplying L'Hopital's Rule x--->8 ln(x + 2) ---> apply 1/(x + 2) lim 1/(x + 2) = 1/(8 + 2) = 1/8 = 0 x--->8 (1/2) * ( 0 - 0 ) - ( 0 - ln(2) ) + (1/2) * ( 0 - ln(3) ) 0 + ln(2) - (1/2) ln(3) ln(2) - ln(3^(1/2)) ln( 2 / v(3)) ------------- ? 1 / [ v(x) * (x + 1) ] dx ---> u-sub u = v(x). . . u^2 = x du = (1/2) * (1/v(x)) dx. . . .2v(x) du = dx ? 1 / [ v(x) * (u^2 + 1) ] 2v(x) du ? 2 / [ (u^2 + 1) ] du . .. . . . . . .. . 1 2 * tan^-1( u ) ] . .. . . . . . . . .0 . .. . . . . . .. .. 1 2 * tan^-1( v(x) ) ] . .. . . . . . . . .. 0 2 * ( tan^-1(v(1)) - tan^-1(v(0)) ) 2 * ( tan^-1(1) - 0 ) 2 * tan^-1(1) 2 * p/4 p/2
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