410 Higher Engineering Mathematics
By dividing out (since the numerator and denominator are of the same degree) and resolving into partial
fractions it was shown in Problem 3, page 14:
x 2 + 1
x 2 − 3x + 2
≡ 1 −
2
(x − 1)
+
5
(x − 2)
Hence
x 2 + 1
x 2 − 3x + 2
dx
≡
1 −
2
(x − 1)
+
5
(x − 2)
dx
= (x −2) ln(x − 1)+ 5 ln(x −2) + c
or x + ln
(x −2)
5
(x −1) 2
+ c
Problem 4. Evaluate
3
2
x 3 − 2x 2 − 4x − 4
x 2 + x − 2
dx,
correct to 4 significant figures.
By dividing out and resolving into partial fractions it
was shown in Problem 4, page 15:
x 3 − 2x 2 − 4x − 4
x 2 + x − 2
≡ x − 3 +
4
(x + 2)
−
3
(x − 1)
Hence
3
2
x 3 − 2x 2 − 4x − 4
x 2 + x − 2
dx
≡
3
2
x − 3 +
4
(x + 2)
−
3
(x − 1)
dx
=
x 2
2
− 3x + 4 ln(x + 2) − 3 ln(x − 1)
3
2
=
9
2
− 9 + 4 ln5 − 3 ln2
− (2 − 6 + 4 ln4 − 3 ln1)
= −1.687, correct to 4 significant figures.
Now try the following exercise
Exercise 162 Further problems on
integration using partial fractions with
linear factors
In Problems 1 to 5, integrate with respect to x.
1.
12
(x 2 − 9)
dx
⎡
⎢
⎣
2 ln(x − 3) − 2 ln(x + 3) + c
or ln
x − 3
x + 3
2
+ c
⎤
⎥
⎦
2.
4(x − 4)
(x 2 − 2x − 3)
dx
⎡
⎢
⎢
⎣
5 ln(x + 1) − ln(x − 3) + c
or ln
(x + 1)
5
(x − 3)
+ c
⎤
⎥
⎥
⎦
3.
3(2x 2 − 8x − 1)
(x + 4)(x + 1)(2x − 1)
dx
⎡
⎢
⎢
⎢
⎢
⎣
7 ln(x + 4) − 3 ln(x + 1)
− ln(2x − 1) + c or
ln
(x + 4) 7
(x + 1) 3 (2x − 1)
+ c
⎤
⎥
⎥
⎥
⎥
⎦
4.
x 2 + 9x + 8
x 2 + x − 6
dx
x + 2 ln(x + 3) + 6 ln(x − 2) + c
or x + ln{(x + 3)
2
(x − 2)
6
} + c
5.
3x 3 − 2x 2 − 16x + 20
(x − 2)(x + 2)
dx
⎡
⎣
3x 2
2
− 2x + ln(x − 2)
−5 ln(x + 2) + c
⎤
⎦
In Problems 6 and 7, evaluate the definite integrals
correct to 4 significant figures.
6.
4
3
x 2 − 3x + 6
x(x − 2)(x − 1)
dx
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