Integration usingpartial fractions 411
7.
∫ 6
4
x^2 −x− 14
x^2 − 2 x− 3
dx [0.8122]
- Determine the value of k, given that:
∫ 1
0
(x−k)
( 3 x+ 1 )(x+ 1 )
dx= 0
[
1
3
]
- The velocity constantkof a given chemical
reaction is given by:
kt=
∫ (
1
( 3 − 0. 4 x)( 2 − 0. 6 x)
)
dx
wherex=0whent=0. Show that:
kt=ln
{
2 ( 3 − 0. 4 x)
3 ( 2 − 0. 6 x)
}
41.3 Worked problemson
integration using partial
fractions with repeated linear
factors
Problem 5. Determine
∫
2 x+ 3
(x− 2 )^2
dx.
It was shown in Problem 5, page 16:
2 x+ 3
(x− 2 )^2
≡
2
(x− 2 )
+
7
(x− 2 )^2
Thus
∫
2 x+ 3
(x− 2 )^2
dx≡
∫ {
2
(x− 2 )
+
7
(x− 2 )^2
}
dx
=2ln(x−2)−
7
(x−2)
+c
⎡
⎣
∫
7
(x− 2 )^2
dxis determined using the algebraic
substitutionu=(x− 2 )— see Chapter 39.
⎤
⎦
Problem 6. Find
∫
5 x^2 − 2 x− 19
(x+ 3 )(x− 1 )^2
dx.
It was shown in Problem 6, page 16:
5 x^2 − 2 x− 19
(x+ 3 )(x− 1 )^2
≡
2
(x+ 3 )
+
3
(x− 1 )
−
4
(x− 1 )^2
Hence
∫
5 x^2 − 2 x− 19
(x+ 3 )(x− 1 )^2
dx
≡
∫ {
2
(x+ 3 )
+
3
(x− 1 )
−
4
(x− 1 )^2
}
dx
=2ln(x+3)+3ln(x−1)+
4
(x−1)
+c
orln
{
(x+3)^2 (x−1)^3
}
+
4
(x−1)
+c
Problem 7. Evaluate
∫ 1
− 2
3 x^2 + 16 x+ 15
(x+ 3 )^3
dx,
correct to 4 significant figures.
It was shown in Problem 7, page 17:
3 x^2 + 16 x+ 15
(x+ 3 )^3
≡
3
(x+ 3 )
−
2
(x+ 3 )^2
−
6
(x+ 3 )^3
Hence
∫
3 x^2 + 16 x+ 15
(x+ 3 )^3
dx
≡
∫ 1
− 2
{
3
(x+ 3 )
−
2
(x+ 3 )^2
−
6
(x+ 3 )^3
}
dx
=
[
3ln(x+ 3 )+
2
(x+ 3 )
+
3
(x+ 3 )^2
] 1
− 2
=
(
3ln4+
2
4
+
3
16
)
−
(
3ln1+
2
1
+
3
1
)
=− 0. 1536 ,correct to 4 significant figures.