Applications of Integrations
Integration by partial decomposition (Examples)
x+ 4 A B
(a)∫ dx=∫ + dx
2
2 x + x−1 2 x−1 x +1
x+4 A B
(i) = +
2 x + x−1 2 x−1 x +1
2
x +4= A ( x +1 ) +B (2 x −1)
x +4=x ( A +2 B ) +(A−B)
(ii) A+2 B=1
A=1−2 B … equation 1
A−B=4 … equation 2
¿ equation 1into equation 2
( 1−2 B )−B=4
−3 B=3
B=−1
¿ B=−1 into equation 1
A=1−2(−1)
A=3
x+ 4 −1 3
Therefore ,∫ dx=∫ + dx
2
2 x + x−1 2 x−1 x+1
1 1
¿−∫ dx+3 ∫ dx
2 x−1 x+1
1
¿− ln ( 2 x−1 )+3 ln ( x +1 ) +c
2
2 x2 +2 x+ 3 A Bx+C
(b)∫ dx=∫ + 2 dx
3
x +3 x x x +3
2 x 2 +2 x +3
= A ( x +3 ) + x ( Bx+C )
2
(i) 3
x +3x
2 x2 +2 x+ 3=x 2 ( A+ B )+Cx +3 A
A+ B=2… equation 1
C=2 … equation 2
Integration by partial decomposition (Examples)
x+ 4 A B
(a)∫ dx=∫ + dx
2
2 x + x−1 2 x−1 x +1
x+4 A B
(i) = +
2 x + x−1 2 x−1 x +1
2
x +4= A ( x +1 ) +B (2 x −1)
x +4=x ( A +2 B ) +(A−B)
(ii) A+2 B=1
A=1−2 B … equation 1
A−B=4 … equation 2
¿ equation 1into equation 2
( 1−2 B )−B=4
−3 B=3
B=−1
¿ B=−1 into equation 1
A=1−2(−1)
A=3
x+ 4 −1 3
Therefore ,∫ dx=∫ + dx
2
2 x + x−1 2 x−1 x+1
1 1
¿−∫ dx+3 ∫ dx
2 x−1 x+1
1
¿− ln ( 2 x−1 )+3 ln ( x +1 ) +c
2
2 x2 +2 x+ 3 A Bx+C
(b)∫ dx=∫ + 2 dx
3
x +3 x x x +3
2 x 2 +2 x +3
= A ( x +3 ) + x ( Bx+C )
2
(i) 3
x +3x
2 x2 +2 x+ 3=x 2 ( A+ B )+Cx +3 A
A+ B=2… equation 1
C=2 … equation 2
