Int

Uitgewerkte oefeningen Hoofdstuk 2
October 1, 2014
Oefening 9
d)
Z
Via Horner bekomen we -1
1
dx
x3 + x2 + x + 1
1
..
.
1
1
1
1
-1
0
0
1
-1
||0
⇒ (x3 + x2 + x + 1) = (x + 1)(x2 + 1)
Splitsen in partieelbreuken:
A
1
Bx + C
=
+ 2
x3 + x2 + x + 1
x+1
x +1
⇔ 1 = A(x2 + 1) + (Bx + C)(x + 1)
⇔ 1 = Ax2 + A + Bx2 + Cx + Bx + C
⇔ 1 = (A + B)x2 + (B + C)x + (A + C)
⇒ (A + B) = 0; (B + C) = 0 en A + C = 1
1
−1
⇒ A = C = en B =
2
2
Z
Z
Z
1
dx
1
1−x
1
dx =
+
dx
x3 + x2 + x + 1
2
x+1 2
x2 + 1
Z
Z
Z
1
dx
1
dx
1
x
=
+
−
dx
2
2
2
x+1 2
x +1 2
x +1
(Stel u = x2 + 1 ⇒ du = 2xdx)
Z
1
1
du
= ln|x + 1| + Bgtan(x) −
2
4
u
1
1
= ln|x + 1| + Bgtan(x) − ln|u| + C
2
4
1
1
= ln|x + 1| + Bgtan(x) − ln|x2 + 1| + C
2
4
1
Oefening 11
r)
Z
√
x
√
dx
x+1+ 3x+1
Stel t6 = x + 1 ⇒ dx = 6t5 dt.
Z
Z 6
x
t −1 5
√
√
dx
=
6t dt
t3 + t2
x+1+ 3x+1
Z 6
(t − 1)t3
=6
dt
t+1
Via Euclidische deling bekomen we:
Z
= 6 t3 (t5 − t4 + t3 − t2 + t − 1)dt
Z
= 6 (t8 − t7 + t6 − t5 + t4 − t3 )dt
6
3
2 9 3 8 6 7
t − t + t − t6 + t5 − t4 + C
3
4
7
5
2
p
p
√
2
3
6
= (x + 1) x + 1 − (x + 1) 3 (x + 1) + (x + 1) 6 (x + 1) − (x + 1)
3
4
7
p
6p
3
+ 6 (x + 1)5 − 3 (x + 1)2 + C
5
2
=
Oefening 12
h)
Z
p
(4 − x2 ) 4 − x2 dx
2
Stel x = 2 sin θ ⇒ dx = 2 cos θdθ en bijgevolg 4−x2 = 4−4 sin2 θ = 4 cos2 θ
Z
Z
p
(4 − x2 ) 4 − x2 dx = 4 cos2 (θ)2 cos(θ)2 cos(θ)dθ
2
1 + cos(2θ)
dθ
2
Z
16
=
1 + 2 cos(2θ) + cos2 (2θ) dθ
4
Z
Z
Z 1 + cos(4θ)
dθ
= 4 dθ + 8 cos 2θdθ + 4
2
1
= 4θ + 4 sin(2θ) + 2θ + sin(4θ) + C
2
= 6θ + 4 sin(2θ) + sin(2θ) cos(2θ) + C
Z
= 16
cos4 (θ)dθ = 16
Z = 6θ + sin(2θ)(4 + cos(2θ)) + C
= 6θ + 2 sin(θ) cos(θ)(4 + 1 − 2 sin2 (θ)) + C
= 6θ + 10 sin(θ) cos(θ) − 4 sin3 (θ) cos(θ) + C
r
r
x2
x2
x
x3
x
1−
1−
−4
+C
= 6Bgsin + 10
2
2
4
8
4
5
x3 p
x
x−
4 − x2 + C
= 6Bgsin +
2
2
4
3

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