3.2 Integration by Substitution


3.2 Integration by Substitution
It is given that ( a x + b ) n d x , n 1.  

(A) Using the Substitution method,
Let u=ax+b Thus,  du dx =a    dx= du a

Example 1:
( 3x+5 ) 3 dx. Let u=3x+5     du dx =3 dx= du 3 ( 3x+5 ) 3 dx = u 3 du 3    substitute 3x+5=u and dx= du 3 = 1 3 u 3 du = 1 3 ( u 4 4 )+c = 1 3 ( ( 3x+5 ) 4 4 )+c    substitute back  u=3x+5   = ( 3x+5 ) 4 12 +c



(B) Using Formula method

( a x + b ) n = ( a x + b ) n + 1 ( n + 1 ) a + c Hence, ( 3 x + 5 ) 3 d x = ( 3 x + 5 ) 4 4 ( 3 ) + c = ( 3 x + 5 ) 4 12 + c

Example 2 (Formula method):

  ( a x + b ) n = ( a x + b ) n + 1 ( n + 1 ) a + c Hence, ( 3 x + 5 ) 3 d x = ( 3 x + 5 ) 4 4 ( 3 ) + c = ( 3 x + 5 ) 4 12 + c

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