Page 31 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
1  1 1 1 1 1 
Hence  2 dx     .  .  dx
( x x  1)  x 2 (x  1) 2 (x 1) 

1 1
 log x  log(x  1)  log(x  1)
2 2
 log x x  1 x  1
 log(x x 2  1)

x
Example: Evaluate  dx
(x  1)(x  1)
Solution:

x A B C
Let  dx    2
(x  1)(x  1) (x  1) (x  2) (x 2)
  A (x  2) 2  (x  1)(x  2)  (x  1)
B
C
x
x  A (x 2  4x  4)  (x 2  3x  2) 1(x  1)
B

comparing the coefficients of x both the sides
A B  0 ....(1)

4A  3B  2C  1 ....(2)

4A  2B C  0 ....(3)
on solving equation (1), (2), (3), we get
A = – 1, B =1, C = 1

dx 1 1
Hence  2    dx   2 dx
(x  1)(x  2) x  1 (x  2)

1
 log(x  1) log(x  2) 

(x  2)
(x  2) 1
 log  Ans .
(x  1) (x  2)

x
Example: Evaluate  dx
2
(x  1)(x  4)
Solution:

x A Bx C

Let 2   2
(x  1)(x  4) x  1 x  4
2
Or x  A (x  4) (Bx C )(x  1)


4
2

x  A (x  4) B (x  x ) (x  1)
C

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