JUST THE MATHS SLIDES NUMBER 12.6 INTEGRATION 6 (Integration by - PDF document

“JUST THE MATHS” SLIDES NUMBER 12.6 INTEGRATION 6 (Integration by partial fractions) by A.J.Hobson 12.6.1 Introduction and illustrations

UNIT 12.6 - INTEGRATION 6 INTEGRATION BY PARTIAL FRACTIONS 12.6.1 INTRODUCTION AND ILLUSTRATIONS The following results will cover most elementary problems involving partial fractions: RESULTS 1. ax + b d x = 1 1 � a ln( ax + b ) + C. 2. a. ( ax + b ) − n +1 ( ax + b ) n d x = 1 1 � + C provided n � = 1 . − n + 1 3. a 2 + x 2 d x = 1 1 a tan − 1 x � a + C. 4. a 2 − x 2 d x = 1 1  a + x    + C when | x | < a, � 2 a ln   a − x and x 2 − a 2 d x = 1 1  x + a   �  + C when | x | > a. 2 a ln   x − a 1

Alternatively, if hyperbolic functions have been stud- ied, a 2 − x 2 d x = 1 1 a tanh − 1 x � a + C. 5. 2 ax + b ax 2 + bx + c d x = ln( ax 2 + bx + c ) + C. � ILLUSTRATIONS We use some of the results of examples on partial fractions in Unit 1.8 1. 7 x + 8 1 3    d x � � (2 x + 3)( x − 1) d x = 2 x + 3 +    x − 1 = 1 2 ln(2 x + 3) + 3 ln( x − 1) + C. 2. 3 x 2 + 9 2 x + 1   � 8 � 8  d x ( x − 5)( x 2 + 2 x + 7) d x = x − 5 +   x 2 + 2 x + 7 6 6  8  2 ln( x − 5) + 1   2 ln( x 2 + 2 x + 7) = 6 ≃ 2 . 427    2

3. 9   − 1 3 1  � � ( x + 1) 2 ( x − 2) = x + 1 − ( x + 1) 2 +  d x     x − 2 3 = − ln( x + 1) + x + 1 + ln( x − 2) + C. 4. 4 x 2 + x + 6 2 2 x + 1    d x. � � ( x − 4)( x 2 + 4 x + 5) d x = x − 4 +   x 2 + 4 x + 5  The second partial fraction has a numerator of 2 x + 1 which is not the derivative of x 2 + 4 x + 5; but we simply rearrange as (2 x + 4) − 3 2 x + 4 3 x 2 + 4 x + 5 ≡ x 2 + 4 x + 5 − ( x + 2) 2 + 1 . By Unit 12.3, Answer = 2 ln( x − 4)+ln( x 2 +4 x +5) − 3tan − 1 ( x +2)+ C. 3