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3.2 Integration by Substitution


3.2 Integration by Substitution
It is given that (ax+b)ndx,n1.  

(A) Using the Substitution method,
Let u=ax+bThus, dudx=a   dx=dua

Example 1:
(3x+5)3dx.Let u=3x+5   dudx=3dx=du3(3x+5)3dx=u3du3  substitute 3x+5=uand dx=du3=13u3du=13(u44)+c=13((3x+5)44)+c   substitute back u=3x+5  =(3x+5)412+c



(B) Using Formula method

(ax+b)n=(ax+b)n+1(n+1)a+cHence,(3x+5)3dx=(3x+5)44(3)+c=(3x+5)412+c

Example 2 (Formula method):

  (ax+b)n=(ax+b)n+1(n+1)a+cHence,(3x+5)3dx=(3x+5)44(3)+c=(3x+5)412+c

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