Tuyển tập chuyên đề và kỹ thuật tích phân, nguyên hàm
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A
1
=
dx
(x + 1) (x + 2)
1
(x + 1) (x + 2)
=
α
x + 1
+
β
x + 2
α , β
α = lim
x→−1
x + 1
(x + 1) (x + 2)
= lim
x→−1
1
x + 2
=
1
−1 + 2
= 1
β = lim
x→−2
x + 2
(x + 1) (x + 2)
= lim
x→−2
1
x + 1
=
1
−2 + 1
= −1
x = 0
1
(0 + 1) (0 + 2)
=
α
0 + 1
+
β
0 + 2
⇔
1
2
= α +
1
2
β
x = 1
1
(1 + 1) (1 + 2)
=
α
1 + 1
+
β
1 + 2
⇔
1
6
=
1
2
α +
1
3
β
α , β
α +
1
2
β =
1
2
1
2
α +
1
3
β =
1
6
⇔
α = 1
β = −1
1
(x + 1) (x + 2)
=
α
x + 1
+
β
x + 2
⇔
1
(x + 1) (x + 2)
=
α (x + 2) + β (x + 1)
(x + 1) (x + 2)
=
x (α + β) + 2α + β
(x + 1) (x + 2)
α + β = 0
2α + β = 1
⇔
α = 1
β = −1
1
(x + 1) (x + 2)
=
1
x + 1
−
1
x + 2
dx
(x + 1) (x + 2)
=
dx
x + 1
−
dx
x + 2
= ln |x + 1| − ln |x + 2| + c = ln
x + 1
x + 2
+ c
A
2
=
x + 2
x (x − 2) (x + 5)
dx
x + 2
x (x − 2) (x + 5)
=
α
x
+
β
x − 2
+
χ
x + 5
α = lim
x→0
x (x + 2)
x (x − 2) (x + 5)
= lim
x→0
x + 2
(x − 2) (x + 5)
=
0 + 2
(0 − 2) (0 + 5)
= −
1
5
β = lim
x→2
(x − 2) (x + 2)
x (x − 2) (x + 5)
= lim
x→2
x + 2
x (x + 5)
=
2 + 2
2 (2 + 5)
=
2
7
χ = lim
x→−5
(x + 5) (x + 2)
x (x − 2) (x + 5)
= lim
x→−5
x + 2
x (x − 2)
=
−5 + 2
−5 (−5 − 2)
= −
3
35
x = −1
−1 + 2
−1 (−1 − 2) (−1 + 5)
=
α
−1
+
β
−1 − 2
+
χ
−1 + 5
⇔ −α −
1
3
β +
1
4
χ =
1
12
x = 1
1 + 2
1 (1 − 2) (1 + 5)
=
α
1
+
β
1 − 2
+
χ
1 + 5
⇔ α − β +
1
6
χ = −
1
2
x = 3
3 + 2
3 (3 − 2) (3 + 5)
=
α
3
+
β
3 − 2
+
χ
3 + 5
⇔
1
3
α + β +
1
8
χ =
5
24
α , β , χ
−α −
1
3
β +
1
4
χ =
1
12
α − β +
1
6
χ = −
1
2
1
3
α + β +
1
8
χ =
5
24
⇔
α = −
1
5
β =
2
7
χ = −
3
35
x + 2
x (x − 2) (x + 5)
=
α
x
+
β
x − 2
+
χ
x + 5
⇔
x + 2
x (x − 2) (x + 5)
=
α (x − 2) (x + 5) + βx (x + 5) + χx (x − 2)
x (x − 2) (x + 5)
⇔
x + 2
x (x − 2) (x + 5)
=
x
2
(α + β + χ) + x (3α + 5β − 2χ) − 10α
x (x − 2) (x + 5)
α + β + χ = 0
3α + 5β − 2χ = 1
−10α = 2
⇔
α = −
1
5
β =
2
7
χ = −
3
35
x + 2
x (x − 2) (x + 5)
= −
1
5x
+
2
7 (x − 2)
−
3
35 (x + 5)
x + 2
x (x − 2) (x + 5)
dx = −
1
5
dx
x
+
2
7
dx
x − 2
−
3
35
dx
x + 5
= −
1
5
ln |x| +
2
7
ln |x − 2| −
3
35
ln |x + 5| + c
A
3
=
x
2
(−3x
2
− 2x + 5) (x + 1)
dx
ax
2
+ bx + c x
1
, , x
2
ax
2
+ bx + c = a (x − x
1
) (x − x
2
)
−3x
2
− 2x + 5 = −3 (x − 1)
x +
5
3
x
2
(−3x
2
− 2x + 5) (x + 1)
= −
x
2
3 (x − 1)
x +
5
3
(x + 1)
=
α
x − 1
+
β
x +
5
3
+
χ
x + 1
α = lim
x→1
−
x
2
(x − 1)
3 (x − 1)
x +
5
3
(x + 1)
= lim
x→1
−
x
2
3
x +
5
3
(x + 1)
= −
1
16
β = lim
x→−
5
3
−
x
2
x +
5
3
3 (x − 1)
x +
5
3
(x + 1)
= lim
x→−
5
3
−
x
2
3 (x − 1) (x + 1)
= −
25
48
χ = lim
x→−1
−
x
2
(x + 1)
3 (x − 1)
x +
5
3
(x + 1)
= lim
x→−1
−
x
2
3 (x − 1)
x +
5
3
=
1
4
x = 0
x
2
(−3x
2
− 2x + 5) (x + 1)
=
α
x − 1
+
β
x +
5
3
+
χ
x + 1
⇔ −α +
3
5
β + χ = 0
x = 2
x
2
(−3x
2
− 2x + 5) (x + 1)
=
α
x − 1
+
β
x +
5
3
+
χ
x + 1
⇔ −
4
33
= α +
3
11
β +
1
3
χ
x = 3
x
2
(−3x
2
− 2x + 5) (x + 1)
=
α
x − 1
+
β
x +
5
3
+
χ
x + 1
⇔ −
9
112
=
1
2
α +
3
14
β +
1
4
χ
−α +
3
5
β + χ = 0
α +
3
11
β +
1
3
χ = −
4
33
1
2
α +
3
14
β +
1
4
χ = −
9
112
⇔
α = −
1
16
β = −
25
48
χ =
1
4
x
2
(−3x
2
− 2x + 5) (x + 1)
= −
x
2
3 (x − 1)
x +
5
3
(x + 1)
=
α
x − 1
+
β
x +
5
3
+
χ
x + 1
⇔ −
x
2
3 (x − 1)
x +
5
3
(x + 1)
=
α
x +
5
3
(x + 1) + β (x − 1) (x + 1) + χ (x − 1)
x +
5
3
(x − 1)
x +
5
3
(x + 1)
⇔ x
2
= x
2
(α + β + χ) + x
8
3
α +
2
3
χ
+
5
3
α − β −
5
3
χ
α + β + χ = 1
8
3
α +
2
3
χ = 0
5
3
α − β −
5
3
χ = 0
⇔
α = −
1
16
β = −
25
48
χ =
1
4
x
2
(−3x
2
− 2x + 5) (x + 1)
= −
x
2
3 (x − 1)
x +
5
3
(x + 1)
= −
1
16 (x − 1)
−
25
48
x +
5
3
+
1
4 (x + 1)
x
2
(−3x
2
− 2x + 5) (x + 1)
dx = −
1
16
dx
x − 1
−
25
48
dx
x +
5
3
+
1
4
dx
x + 4
= −
1
16
ln |x − 1| −
25
48
ln
x +
5
3
+
1
4
ln |x + 4| + c
A
4
=
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
dx
ax
2
+ bx + c = 0 ∆ = b
2
− 4ac < 0
ax
2
+ bx + c = a (x − x
1
) (x − x
2
) x
1
= α + βi, x
2
= α − βi, i
2
= −1
x
2
+ 4x + 5 = (x + 2 + i) (x + 2 − i)
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
x − 1
(x + 2 + i) (x + 2 − i) (x
2
− 4)
=
α
x + 2 + i
+
β
x + 2 − i
+
χ
x − 2
+
δ
x + 2
α = lim
x→−2−i
(x − 1) (x + 2 + i)
(x + 2 + i) (x + 2 − i) (x
2
− 4)
= lim
x→−2−i
x − 1
(x + 2 − i) (x
2
− 4)
= −
13
34
−
1
34
i
β = lim
x→−2+i
(x − 1) (x + 2 − i)
(x + 2 + i) (x + 2 − i) (x
2
− 4)
= lim
x→−2+i
x − 1
(x + 2 + i) (x
2
− 4)
= −
13
34
+
1
34
i
χ = lim
x→2
(x − 1) (x − 2)
(x
2
+ 4x + 5) (x − 2) (x + 2)
= lim
x→2
x − 1
(x
2
+ 4x + 5) (x + 2)
=
1
68
δ = lim
x→−2
(x − 1) (x + 2)
(x
2
+ 4x + 5) (x − 2) (x + 2)
= lim
x→−2
x − 1
(x
2
+ 4x + 5) (x − 2)
=
3
4
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
−
13
34
−
1
34
i
x + 2 + i
+
−
13
34
+
1
34
i
x + 2 − i
+
1
68
x − 2
+
3
4
x + 2
=
−13x − 27
17 (x
2
+ 4x + 5)
+
1
68 (x − 2)
+
3
4 (x + 2)
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
αx + β
x
2
+ 4x + 5
+
χ
x − 2
+
δ
x + 2
x = 0
1
5
β −
1
2
χ +
1
2
δ =
1
20
x = 1
1
10
α +
1
10
β − χ +
1
3
δ = 0
x = 3
3
26
α +
1
26
β + χ +
1
5
δ =
1
65
x = 4
4
37
α +
1
37
β +
1
2
χ +
1
6
δ =
1
148
1
5
β −
1
2
χ +
1
2
δ =
1
20
1
10
α +
1
10
β − χ +
1
3
δ = 0
3
26
α +
1
26
β + χ +
1
5
δ =
1
65
4
37
α +
1
37
β +
1
2
χ +
1
6
δ =
1
148
⇔
δ = −
2
5
β + χ +
1
10
1
10
α −
1
30
β −
2
3
χ = −
1
30
3
26
α −
27
650
β +
6
5
χ = −
3
650
4
37
α −
22
555
β +
2
3
χ = −
11
1110
⇔
α = −
13
17
β = −
27
17
χ =
1
68
δ =
3
4
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
−13x − 27β
17 (x
2
+ 4x + 5)
+
χ
68 (x − 2)
+
3δ
4 (x + 2)
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
αx + β
x
2
+ 4x + 5
+
χ
x − 2
+
δ
x + 2
⇔ x − 1 = (αx + β)
x
2
− 4
+ χ
x
2
+ 4x + 5
(x + 2) + δ
x
2
+ 4x + 5
(x − 2)
⇔ x − 1 = x
3
(α + χ + δ) + x
2
(β + 6χ + 2δ) + x (−4α + 13χ − 3δ) + (−4β + 10χ − 10δ)
α + χ + δ = 0
β + 6χ + 2δ = 0
−4α + 13χ − 3δ = 1
4β − 10χ + 10δ = 1
⇔
δ = −α −χ
β + 6χ + 2 (−α − χ) = 0
−4α + 13χ − 3 (−α − χ) = 1
4β − 10χ + 10 (−α − χ) = 1
⇔
δ = −α −χ
−2α + β + 4χ = 0
−α + 16χ = 1
−10α + 4β − 20χ = 1
⇔
α = −
13
17
β = −
27
17
χ =
1
68
δ =
3
4
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
=
−13x − 27
17 (x
2
+ 4x + 5)
+
1
68 (x − 2)
+
3
4 (x + 2)
x − 1
(x
2
+ 4x + 5) (x
2
− 4)
dx = −
13
34
2x +
54
13
x
2
+ 4x + 5
dx +
1
68
dx
x − 2
+
3
4
dx
x + 2
= −
13
34
(2x + 4)
x
2
+ 4x + 5
dx −
1
17
dx
x
2
+ 4x + 5
+
1
68
dx
x − 2
+
3
4
dx
x + 2
= −
13
34
d
x
2
+ 4x + 5
x
2
+ 4x + 5
−
1
17
dx
(x + 2)
2
+ 1
+
1
68
dx
x − 2
+
3
4
dx
x + 2
= −
13
34
ln
x
2
+ 4x + 5
−
1
17
arctan (x + 2) +
1
68
ln |x − 2| +
3
4
ln |x + 2| + c
A
5
=
x
x
3
+ 1
dx
x
x
3
+ 1
=
x
(x + 1) (x
2
− x + 1)
=
α
x + 1
+
β
x −
1
2
−
√
3
2
i
+
χ
x −
1
2
+
√
3
2
i
α = lim
x→−1
x
x
2
− x + 1
= −
1
3
β = lim
x→
1
2
+
√
3
2
i
x
(x + 1)
x −
1
2
+
√
3
2
i
=
1
6
−
√
3
6
i
χ = lim
x→
1
2
−
√
3
2
i
x
(x + 1)
x −
1
2
−
√
3
2
i
=
1
6
+
√
3
6
i
x
x
3
+ 1
= −
1
3 (x + 1)
+
1
6
−
√
3
6
i
x −
1
2
−
√
3
2
i
+
1
6
+
√
3
6
i
x −
1
2
+
√
3
2
i
= −
1
3 (x + 1)
+
x + 1
3 (x
2
− x + 1)
x
x
3
+ 1
=
α
x + 1
+
βx + χ
x
2
− x + 1
x = 0 0 = α + χ
x = 1
1
2
=
1
2
α + β + χ
x = 2
2
9
=
1
3
α +
2
3
β +
1
3
χ
α + χ = 0
1
2
α + β + χ =
1
2
1
3
α +
2
3
β +
1
3
χ =
2
9
⇔
α = −
1
3
β =
1
3
χ =
1
3
x
x
3
+ 1
= −
1
3 (x + 1)
+
x + 1
3 (x
2
− x + 1)
x
x
3
+ 1
=
α
x + 1
+
βx + χ
x
2
− x + 1
⇔ x = α
x
2
− x + 1
+ (βx + χ) (x + 1)
α + β = 0
−α + β + χ = 1
α + χ = 0
⇔
α = −
1
3
β =
1
3
χ =
1
3
x
x
3
+ 1
dx = −
1
3
dx
x + 1
+
1
6
2x + 2
x
2
− x + 1
dx
= −
1
3
dx
x + 1
+
1
6
2x + 1
x
2
− x + 1
dx +
1
6
dx
x
2
− x + 1
= −
1
3
dx
x + 1
+
1
6
d
x
2
− x + 1
x
2
− x + 1
+
1
6
dx
x −
1
2
2
+
3
4
= −
1
3
ln |x + 1| +
1
6
ln
x
2
− x + 1
+
1
3
√
3
arctan
2x − 1
√
3
+ c
A
6
=
dx
x
8
+ 1
x
8
+ 1 =
x
4
+ px
2
+ 1
x
4
− px
2
+ 1
= x
8
+
2 − p
2
x
4
+ 1
2 − p
2
= 0 ⇔ p = ±
√
2
⇒ x
8
+ 1 =
x
4
+
√
2x
2
+ 1
x
4
−
√
2x
2
+ 1
x
4
+
√
2x
2
+ 1 =
x
2
+ qx + 1
x
2
− qx + 1
= x
4
+
2 − q
2
x
2
+ 1
2 − q
2
=
√
2 ⇔ q = ±
2 −
√
2
⇒ x
4
+
√
2x
2
+ 1 =
x
2
+
2 −
√
2x + 1
x
2
−
2 −
√
2x + 1
x
4
−
√
2x
2
+ 1 =
x
2
+ rx + 1
x
2
− rx + 1
= x
4
+
2 − r
2
x
2
+ 1
2 − r
2
= −
√
2 ⇔ r = ±
2 +
√
2
⇒ x
4
−
√
2x
2
+ 1 =
x
2
+
2 +
√
2x + 1
x
2
−
2 +
√
2x + 1
x
8
+ 1 =
x
2
+
2 −
√
2x + 1
x
2
−
2 −
√
2x + 1
x
2
+
2 +
√
2x + 1
x
2
−
2 +
√
2x + 1
⇒
1
x
8
+ 1
=
1
8
2 −
√
2x + 2
x
2
+
2 −
√
2x + 1
+
1
8
−
2 −
√
2x + 2
x
2
−
2 −
√
2x + 1
+
1
8
2 +
√
2x + 2
x
2
+
2 +
√
2x + 1
+
+
1
8
−
2 +
√
2x + 2
x
2
−
2 +
√
2x + 1
dx
ax
2
+ bx + c
=
1
a
dx
x
2
+
b
a
x +
c
a
, 4ac − b
2
> 0
=
1
a
dx
x −
−b
2a
2
+
4ac − b
2
2a
2
=
1
a
dx
(x − p)
2
+ q
2
(1)
p =
−b
2a
, q =
√
4ac − b
2
2a
, x = p + qt
=
1
aq
1
t
2
+ 1
dt =
2
√
4ac − b
2
1
t
2
+ 1
dt =
2
√
4ac − b
2
arctan t
=
2
√
4ac − b
2
arctan
2ax + b
√
4ac − b
2
+ C
t =
x − p
q
=
2ax + b
√
4ac − b
2
xdx
ax
2
+ bx + c
=
1
2a
2ax + b − b
ax
2
+ bx + c
dx =
1
2a
2ax + b
ax
2
+ bx + c
dx −
b
2a
1
ax
2
+ bx + c
dx
=
1
2a
ln
ax
2
+ bx + c
−
b
2a
dx
ax
2
+ bx + c
+ C
a = c = 1, ∆ = 4 − b
2
> 0
Ax + B
x
2
+ bx + 1
dx = A
xdx
x
2
+ bx + 1
+ B
dx
x
2
+ bx + 1
=
2B −Ab
√
4 − b
2
arctan
2x + b
√
4 − b
2
+
A
2
ln
x
2
+ bx + 1
+ C
A
6.1
=
1
8
2 −
√
2x + 2
x
2
+
2 −
√
2x + 1
dx =
=
2 +
√
2
8
arctan
2x +
2 −
√
2
2 +
√
2
+
2 −
√
2
16
ln
x
2
+
2 −
√
2x + 1
+ C
1
A
6.2
=
1
8
−
2 −
√
2x + 2
x
2
−
2 −
√
2x + 1
dx =
=
2 +
√
2
8
arctan
2x −
2 −
√
2
2 +
√
2
−
2 −
√
2
16
ln
x
2
−
2 −
√
2x + 1
+ C
2
A
6.3
=
2 +
√
2x + 2
x
2
+
2 +
√
2x + 1
dx =
=
2 −
√
2
8
arctan
2x +
2 −
√
2
2 −
√
2
+
2 +
√
2
16
ln
x
2
+
2 +
√
2x + 1
+ C
3
A
6.4
=
−
2 +
√
2x + 2
x
2
−
2 +
√
2x + 1
dx =
=
2 −
√
2
8
arctan
2x −
2 −
√
2
2 −
√
2
−
2 +
√
2
16
ln
x
2
−
2 +
√
2x + 1
+ C
4
A
6
= A
6.1
+ A
6.2
+ A
6.3
+ A
6.4
x
8
+ 1 = 0 ⇔ x
8
= −1 = cos (π + k2π) + i sin (π + k2π)
⇒ x = cos
π + k2π
8
+ i sin
π + k2π
8
k = 0, , 7
sin
π
8
= sin
7π
8
= cos
3π
8
=
1
2
2 −
√
2, sin
3π
8
= sin
5π
8
= cos
π
8
=
1
2
2 +
√
2 ,
cos
5π
8
= cos
7π
8
= −
1
2
2 −
√
2 ,
dx
1 + x
8
= −
1
8
3
k=0
ln
x
2
− 2x cos
(2k + 1) π
8
+ 1
× cos
(2k + 1) π
8
+
+
1
4
3
k=0
arctan
x sin
(2k + 1) π
8
1 − x cos
(2k + 1) π
8
× sin
(2k + 1) π
8
+ C, k = 0, , 7
