3 1 x  dx dx 2 4 x We the integrand into two partial fractions each can be integrated using the extension of the formulae for (ax + b) = 2 ln (x + 3) ln (x 4) + A – – (x 11) Then = ln (x +3)2 ln (x 4) + A – – (x 3)2  = ln 4 x  A 6.4 Integration of expression with repeated linear factors in the denominator. Consider 1 (1) dx (x 2)(x 1)2  The integrand is a proper fraction and with repeated linear factors It is written as 1 A B C (x 2)(x 1)(x 1)    2 (x 1)2 x  x 1)2 2)(x 1) 2)     A(x B(x C(x = (x 2)(x 1)(x 1)  Then 1 = A (x-1)2 + B (x +2) (x 1) + C (x + 2) (2) – Let us put x = 1 on both sides of (1) so that the terms containing A and B will become zero. 1 Then 1 = A (0) + B (0) + C (1+2), Hence C = 3 Similarly put x = -2 on both sides of (1) so that the terms containing B and C become zero. 1 Then 1 = A (-2 1)2 + B (0) + C (0), Hence A = – 9 . To get the value of B either we can give one more value other than 1 and for x or –2 else we can compare the coefficients of x2 on both sides of (2) 0=A+B 1 0= 1 9 + B then B = - . 9 Hence the integral (1) becomes 58 1 1  dx 1 1 2
2 x  3 dx 1 x 4 dx We the integrand into two partial fractions each can be integrated using the extension of the formulae for (ax + b) Then (x 11) = 2 ln (x + 3) – ln (x – 4) + A = ln (x +3)2 – ln (x – 4) + A = ln (x  3)2 x 4  A 6.4 Integration of expression with repeated linear factors in the denominator. Consider 1 (x  2)(x 1)2 dx (1) The integrand is a proper fraction and with repeated linear factors It is written as 1 (x  2)(x 1)(x 1)  A x  2  B x C (x 1)2 = A(x 1)2  B(x  2)(x 1)  C(x  2) (x  2)(x 1)(x 1) Then 1 = A (x-1)2 + B (x +2) (x – 1) + C (x + 2) (2) Let us put x = 1 on both sides of (1) so that the terms containing A and B will become zero. Then 1 = A (0) + B (0) + C (1+2), Hence C = 1 3 Similarly put x = -2 on both sides of (1) so that the terms containing B and C become zero. Then 1 = A (-2 – 1)2 + B (0) + C (0), Hence A = 1 9 . To get the value of B either we can give one more value other than 1 and –2 for x or else we can compare the coefficients of x2 on both sides of (2) 0=A+B 0= 1 9 + B then B = - 1 9 . Hence the integral (1) becomes 58 1 2 dx  1 1 1
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