334
8 Secondary Ventilation
z
3
− 1
= (z − 1)
z
2
+ z + 1
Thus:
−2
z 3 − 1
=
A
z − 1
+
Bz + C
z 2 + z + 1
where
17 :
A = −
2
3
B =
2
3
C =
4
3
In this way, the new integral is:
I (z) = −
2
3
dz
z − 1
+
2
3
z + 2
z 2 + z + 1
dz = I (z 1 ) + I (z 2 )
Therefore:
I (z 1 ) = −
2
3
dz
z − 1
= −
2
3
ln(z − 1) = −
1
3
ln(z − 1)
2
And
18 :
I (z 2 ) =
2
3
z + 2
z 2 + z + 1
dz =
1
3
2z + 1
z 2 + z + 1
dz +
1
3
3
z 2 + z + 1
dz
Thus:
I (z 2 ) =
1
3
ln
z
2
+ z + 1
+
2
√
3
arctan
2z + 1
√
3
The solution to Eq. 8.53, changing the sign of both members, is:
I (z 1 ) + I (z 2 ) =
1
3
ln(z − 1)
2
−
1
3
ln
z
2
+ z + 1
−
2
√
3
arctan
2z + 1
√
3
Therefore, we obtain Eq. 8.54:
17 For the system to be solved: A + B = 0; A – B + C = 0; A − C = −2.
18 The second term can be solved by expressing it in the form 1/(u 2 + 1) which when integrated
gives arctan (u). Programmes such as Symbolab, Matlab and Mathematica, amongst others, will
give an immediate result for the integral of 1/(x 2 + x + 1).