2.5 Proof of the Ten Extension Formulae We shall prove some of he Ten Extension formulae using the substitution u = ax + b so that dx = du Example 1 a (ax  b)n1  a). that (ax   b)n dx  1) a(n  c  b). Hence find (4x 9)20  dx 16 Solution du Let u = a x + b so that dx = a  a) Then (ax   b)n dx undu 1 = a n 1 un1 = 1  a n  c (ax  b)n1 = 1)  a(n  c x21 20 21 ,  c 21.4  ,  c = (4x 9)21  84  c (4x 9)21  (4x  9)20 = dx Example 2 1 ln(ax  b) a). Prove that a  c
2.5 Proof of the Ten Extension Formulae We shall prove some of he Ten Extension formulae using the substitution u = ax + b so that dx = Example 1 du a a). that (ax  b)n dx  (ax  b)n1 a(n  1)  c b). Hence find (4x  9)20 dx 16 Solution Let u = a x + b so that dx = du a a) Then (ax  b)n dx  undu = 1 a n = 1 un1 a n  1  c = (ax  b)n1 a(n  1)  c 20 x21 21  c , (4x  9)20 dx = = (4x  9)21 21.4 (4x  9)21 84  c ,  c Example 2 a). Prove that 1 ln(ax  b) a  c    
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