Page 32 - DMTH202_BASIC_MATHEMATICS_II

Unit 2: Integration by Partial Fraction

comparing the coefficients of x both side Notes
A + B = 0 ....(1)
– B + C = 1 ....(2)
4A – C = 0 ....(3)
adding (1) and (2) we get
A + C = 1 ...(4)
Adding (3) and (4) we get

5A=1 a=1/5

1
 B  
5
6
0rC 
5

x 1 1 1 x  6
Hence  2 dx   dx  2 
(x  1)(x  4) 5 (x  1) 5 x  4

1 1 1 x 6 1
  dx  2  dx   dx
2
5 x  1 5 x  4 5 x  4
1 1 1 6 1 x
2
 log(x  1)  . log(x  4)  tan 
5 5 2 5x 2 2
1 1 3

2
1
 log(x  1)  log(x  4)  tan x /2
5 10 5
4
2
x
x  x   1
Example: Evaluate  3 2 dx
x  x
Solution:
Here the integrand is an improper fraction therefore we divide the numerator by the denominator
and obtain
4
2
x
x  x   1 x  1 x  1
 x   x 
3
x  x 2 x  x 2 x 2 (x  1)
3
How, let
x  1 A B C
  
x 2 (x  1) x x 2 x  1
 x  1  Ax (x  1) B (x  1) Cx 2


on comparing the coefficients of x, we get
A + C = 0 ....(1)
B = 1
–A–B = 1
–A = 2  A = – 2

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