Integration usingpartial fractions 411
7.
6
4
x 2 − x − 14
x 2 − 2x − 3
dx
[0.8122]
8. Determine the value of k, given that:
1
0
(x − k)
(3x + 1)(x + 1)
dx = 0
1
3
9. The velocity constant k of a given chemical
reaction is given by:
kt =
1
(3 − 0.4x)(2 − 0.6x)
dx
where x = 0 when t = 0. Show that:
kt = ln
2(3 − 0.4x)
3(2 − 0.6x)
41.3 Worked problems on
integration using partial
fractions with repeated linear
factors
Problem 5. Determine
2x + 3
(x − 2) 2 dx.
It was shown in Problem 5, page 16:
2x + 3
(x − 2) 2 ≡
2
(x − 2)
+
7
(x − 2) 2
Thus
2x + 3
(x − 2) 2 dx ≡
2
(x − 2)
+
7
(x − 2) 2
dx
= 2 ln(x −2) −
7
(x −2)
+ c
⎡
⎣
7
(x − 2) 2 dx is determined using the algebraic
substitution u = (x − 2) — see Chapter 39.
⎤
⎦
Problem 6. Find
5x
2
− 2x − 19
(x + 3)(x − 1) 2 dx.
It was shown in Problem 6, page 16:
5x 2 − 2x − 19
(x + 3)(x − 1) 2 ≡
2
(x + 3)
+
3
(x − 1)
−
4
(x − 1) 2
Hence
5x 2 − 2x − 19
(x + 3)(x − 1) 2 dx
≡
2
(x + 3)
+
3
(x − 1)
−
4
(x − 1) 2
dx
= 2 ln (x +3) + 3 ln (x −1)+
4
(x − 1)
+ c
or ln
(x +3)
2 (x −1)
3
+
4
(x − 1)
+ c
Problem 7. Evaluate
1
−2
3x 2 + 16x + 15
(x + 3) 3
dx,
correct to 4 significant figures.
It was shown in Problem 7, page 17:
3x 2 + 16x + 15
(x + 3) 3
≡
3
(x + 3)
−
2
(x + 3) 2 −
6
(x + 3) 3
Hence
3x 2 + 16x + 15
(x + 3) 3
dx
≡
1
−2
3
(x + 3)
−
2
(x + 3) 2 −
6
(x + 3) 3
dx
=
3 ln(x + 3) +
2
(x + 3)
+
3
(x + 3) 2
1
−2
=
3 ln4 +
2
4
+
3
16
−
3 ln1 +
2
1
+
3
1
= −0.1536, correct to 4 significant figures.