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Интегралы, содержащие корни - LibreTimes
August 13, 2026 · Reference
Интегралы, содержащие корни
Sergey
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Неопределённые интегралы от выражений, содержащих корни.
∫
x
−
a
d
x
=
3
2
(
x
−
a
)
3/2
+
C
∫
x
±
a
1
d
x
=
2
x
±
a
+
C
∫
a
−
x
1
d
x
=
−
2
a
−
x
+
C
∫
x
x
−
a
d
x
=
⎩
⎨
⎧
3
2
a
(
x
−
a
)
3/2
+
5
2
(
x
−
a
)
5/2
,
или
3
2
x
(
x
−
a
)
3/2
−
15
4
(
x
−
a
)
5/2
,
или
15
2
(
2
a
+
3
x
)
(
x
−
a
)
3/2
+
C
∫
a
x
+
b
d
x
=
(
3
a
2
b
+
3
2
x
)
a
x
+
b
+
C
∫
(
a
x
+
b
)
3/2
d
x
=
5
a
2
(
a
x
+
b
)
5/2
+
C
∫
x
±
a
x
d
x
=
3
2
(
x
∓
2
a
)
x
±
a
+
C
∫
a
−
x
x
d
x
=
−
x
(
a
−
x
)
−
a
arct
g
x
−
a
x
(
a
−
x
)
+
C
∫
a
+
x
x
d
x
=
x
(
a
+
x
)
−
a
ln
[
x
+
x
+
a
]
+
C
∫
x
a
x
+
b
d
x
=
15
a
2
2
(
−
2
b
2
+
ab
x
+
3
a
2
x
2
)
a
x
+
b
+
C
∫
x
(
a
x
+
b
)
d
x
=
4
a
3/2
1
[
(
2
a
x
+
b
)
a
x
(
a
x
+
b
)
−
b
2
ln
a
x
+
a
(
a
x
+
b
)
]
+
C
∫
x
3
(
a
x
+
b
)
d
x
=
[
12
a
b
−
8
a
2
x
b
2
+
3
x
]
x
3
(
a
x
+
b
)
+
8
a
5/2
b
3
ln
a
x
+
a
(
a
x
+
b
)
+
C
∫
x
2
±
a
2
d
x
=
2
1
x
x
2
±
a
2
±
2
1
a
2
ln
x
+
x
2
±
a
2
+
C
∫
a
2
−
x
2
d
x
=
2
1
x
a
2
−
x
2
+
2
1
a
2
arct
g
a
2
−
x
2
x
+
C
∫
x
x
2
±
a
2
d
x
=
3
1
(
x
2
±
a
2
)
3/2
+
C
∫
x
2
±
a
2
1
d
x
=
ln
x
+
x
2
±
a
2
+
C
∫
a
2
−
x
2
1
d
x
=
arcsin
a
x
+
C
∫
x
2
±
a
2
x
d
x
=
x
2
±
a
2
+
C
∫
a
2
−
x
2
x
d
x
=
−
a
2
−
x
2
+
C
∫
x
2
±
a
2
x
2
d
x
=
2
1
x
x
2
±
a
2
∓
2
1
a
2
ln
x
+
x
2
±
a
2
+
C
∫
a
x
2
+
b
x
+
c
d
x
=
4
a
b
+
2
a
x
a
x
2
+
b
x
+
c
+
8
a
3/2
4
a
c
−
b
2
ln
2
a
x
+
b
+
2
a
(
a
x
2
+
b
x
+
c
)
+
C
∫
x
a
x
2
+
b
x
+
c
d
x
=
48
a
5/2
1
(
2
a
a
x
2
+
b
x
+
c
(
−
3
b
2
+
2
ab
x
+
8
a
(
c
+
a
x
2
)
)
+
3
(
b
3
−
4
ab
c
)
ln
b
+
2
a
x
+
2
a
a
x
2
+
b
x
+
c
)
+
C
∫
a
x
2
+
b
x
+
c
1
d
x
=
a
1
ln
2
a
x
+
b
+
2
a
(
a
x
2
+
b
x
+
c
)
+
C
∫
a
x
2
+
b
x
+
c
x
d
x
=
a
1
a
x
2
+
b
x
+
c
−
2
a
3/2
b
ln
2
a
x
+
b
+
2
a
(
a
x
2
+
b
x
+
c
)
+
C
∫
(
a
2
+
x
2
)
3/2
d
x
=
a
2
a
2
+
x
2
x
+
C