A binomial integral is an integral of the form:
∫
x
m
(
a
x
n
+
b
)
p
d
x
{\ displaystyle \ int x ^ {m} \ left (ax ^ {n} + b \ right) ^ {p} \, \ mathrm {d} x}
, where are rational numbers and .
m
,
n
,
p
{\ displaystyle m, \ n, \ p}
a
≠
0
,
n
≠
0
{\ displaystyle a \ neq 0, n \ neq 0}
The set of Chebyshev now makes a statement when a binomial integral is elementary integrated. Elementary integrable means that the integral can be determined with the help of an antiderivative.
Chebyshev theorem
statement
A binomial integral can be elementarily integrated if and only if at least one of the numbers is or is whole .
p
,
m
+
1
n
{\ displaystyle \ textstyle p, \ {\ frac {m + 1} {n}}}
m
+
1
n
+
p
{\ displaystyle \ textstyle {\ frac {m + 1} {n}} + p}
If the function can be integrated elementarily, the antiderivative can be determined in the following three cases :
p
∈
Z
{\ displaystyle p \ in \ mathbb {Z}}
with the substitution where q is the main denominator of m and n
x
=
t
q
{\ displaystyle x = t ^ {q}}
m
+
1
n
∈
Z
{\ displaystyle {\ frac {m + 1} {n}} \ in \ mathbb {Z}}
with the substitution where q is the denominator of p
t
q
=
a
x
n
+
b
{\ displaystyle t ^ {q} = ax ^ {n} + b}
m
+
1
n
+
p
∈
Z
{\ displaystyle {\ frac {m + 1} {n}} + p \ in \ mathbb {Z}}
with the substitution where q is the denominator of p.
t
q
=
a
x
n
+
b
x
n
{\ displaystyle t ^ {q} = {\ frac {ax ^ {n} + b} {x ^ {n}}}}
Examples
1st example
∫
1
1
+
x
4th
d
x
=
∫
x
0
(
1
⋅
x
4th
+
1
)
-
1
2
d
x
⇒
m
=
0
,
n
=
4th
,
p
=
-
1
2
⇒
p
=
-
1
2
∉
Z
,
m
+
1
n
=
0
+
1
4th
=
1
4th
∉
Z
,
m
+
1
n
+
p
=
0
+
1
4th
-
1
2
=
-
1
4th
∉
Z
{\ displaystyle {\ begin {aligned} & \ int {\ frac {1} {\ sqrt {1 + x ^ {4}}}} \, \ mathrm {d} x = \ int x ^ {0} \ left (1 \ cdot x ^ {4} +1 \ right) ^ {- {\ frac {1} {2}}} \, \ mathrm {d} x \\\ Rightarrow & m = 0, \ n = 4, \ p = - {\ frac {1} {2}} \\\ Rightarrow & p = - {\ frac {1} {2}} \ notin \ mathbb {Z}, \ {\ frac {m + 1} {n} } = {\ frac {0 + 1} {4}} = {\ frac {1} {4}} \ notin \ mathbb {Z}, \ {\ frac {m + 1} {n}} + p = { \ frac {0 + 1} {4}} - {\ frac {1} {2}} = - {\ frac {1} {4}} \ notin \ mathbb {Z} \ end {aligned}}}
It is therefore not fundamentally integrable.
1
1
+
x
4th
{\ displaystyle \ textstyle {\ frac {1} {\ sqrt {1 + x ^ {4}}}}}
2nd example
∫
x
3
(
x
+
1
)
2
3
d
x
=
∫
x
1
3
(
1
⋅
x
1
+
1
)
2
3
d
x
⇒
m
=
1
3
,
n
=
1
,
p
=
2
3
⇒
p
=
2
3
∉
Z
,
m
+
1
n
=
1
3
+
1
1
=
4th
3
∉
Z
,
m
+
1
n
+
p
=
1
3
+
1
1
+
2
3
=
2
∈
Z
{\ displaystyle {\ begin {aligned} & \ int {\ sqrt [{3}] {x}} {\ sqrt [{3}] {\ left (x + 1 \ right) ^ {2}}} \, \ mathrm {d} x = \ int x ^ {\ frac {1} {3}} \ left (1 \ cdot x ^ {1} +1 \ right) ^ {\ frac {2} {3}} \, \ mathrm {d} x \\\ Rightarrow & m = {\ frac {1} {3}}, \ n = 1, \ p = {\ frac {2} {3}} \\\ Rightarrow & p = {\ frac {2} {3}} \ notin \ mathbb {Z}, \ {\ frac {m + 1} {n}} = {\ frac {{\ frac {1} {3}} + 1} {1}} = {\ frac {4} {3}} \ notin \ mathbb {Z}, \ {\ frac {m + 1} {n}} + p = {\ frac {{\ frac {1} {3}} + 1} {1}} + {\ frac {2} {3}} = 2 \ in \ mathbb {Z} \ end {aligned}}}
So is elementary integrable.
x
3
(
x
+
1
)
2
3
{\ displaystyle \ textstyle {\ sqrt [{3}] {x}} {\ sqrt [{3}] {\ left (x + 1 \ right) ^ {2}}}}
source
binomial integral . In: Guido Walz (Ed.): Lexicon of Mathematics . 1st edition. Spectrum Academic Publishing House, Mannheim / Heidelberg 2000, ISBN 3-8274-0439-8 .
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