Indefinite integral of reciprocal n degree polynomial
I find that the Torricelli's trumpet has very interesting properties.
It's surface:
$$int_{1}^{infty}frac{2pi}{x}dx=2piln{x}|_{1}^{infty}rightarrowinfty.$$
is infinity.
Its volume:
$$int_{1}^{infty}frac{pi}{x^{2}}dx=-frac{pi}{x}|_{1}^{infty}=pi.$$
is finite.
I extend it so its radius will be:
$$y=frac{1}{sqrt{x^2+c}},c is a constant.$$
Its surface will be:
$$int_{1}^{infty}frac{2pi}{sqrt{x^2+c}}dx=2piln left|frac{x}{sqrt{c}}+sqrt{frac{c+x^2}{c}}right||_1^inftyrightarrow diverges.$$
Then its volume:
$$int_{1}^{infty}frac{pi}{x^2+c}dx=frac{pi ^2}{2sqrt{c}}-frac{pi arctan left(sqrt{frac{1}{c}}right)}{sqrt{c}}rightarrow converges.$$
I tried to extend further for the radius:
$$y=frac{1}{sqrt[3]{x^3+c}}.\$$
Its surface:
$$int_{1}^{infty} frac{2pi}{sqrt[3]{x^3+c}} dx.\$$
Its volume:
$$ int_{1}^{infty} frac{pi}{(x^3+c)^{2/3}} dx.\$$
I can't find a general formula for
$$int_ frac{1}{sqrt[3]{x^3+c}} dx.\$$
If c is polynomial and its degree will be less then the x term before it.
How would I do a general expression for:
$$int_ frac{1}{sqrt[n]{ax^n+bx^{n-1}+cdots+z}} dx.\$$
polynomials soft-question improper-integrals indefinite-integrals
edited yesterday
Harry Peter
5,45111439
asked Dec 24 at 6:58
John He
94
add a comment |
I find that the Torricelli's trumpet has very interesting properties.
It's surface:
$$int_{1}^{infty}frac{2pi}{x}dx=2piln{x}|_{1}^{infty}rightarrowinfty.$$
is infinity.
Its volume:
$$int_{1}^{infty}frac{pi}{x^{2}}dx=-frac{pi}{x}|_{1}^{infty}=pi.$$
is finite.
I extend it so its radius will be:
$$y=frac{1}{sqrt{x^2+c}},c is a constant.$$
Its surface will be:
$$int_{1}^{infty}frac{2pi}{sqrt{x^2+c}}dx=2piln left|frac{x}{sqrt{c}}+sqrt{frac{c+x^2}{c}}right||_1^inftyrightarrow diverges.$$
Then its volume:
$$int_{1}^{infty}frac{pi}{x^2+c}dx=frac{pi ^2}{2sqrt{c}}-frac{pi arctan left(sqrt{frac{1}{c}}right)}{sqrt{c}}rightarrow converges.$$
I tried to extend further for the radius:
$$y=frac{1}{sqrt[3]{x^3+c}}.\$$
Its surface:
$$int_{1}^{infty} frac{2pi}{sqrt[3]{x^3+c}} dx.\$$
Its volume:
$$ int_{1}^{infty} frac{pi}{(x^3+c)^{2/3}} dx.\$$
I can't find a general formula for
$$int_ frac{1}{sqrt[3]{x^3+c}} dx.\$$
If c is polynomial and its degree will be less then the x term before it.
How would I do a general expression for:
$$int_ frac{1}{sqrt[n]{ax^n+bx^{n-1}+cdots+z}} dx.\$$
polynomials soft-question improper-integrals indefinite-integrals
edited yesterday
Harry Peter
5,45111439
asked Dec 24 at 6:58
John He
94
1
It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28
add a comment |
I find that the Torricelli's trumpet has very interesting properties.
It's surface:
$$int_{1}^{infty}frac{2pi}{x}dx=2piln{x}|_{1}^{infty}rightarrowinfty.$$
is infinity.
Its volume:
$$int_{1}^{infty}frac{pi}{x^{2}}dx=-frac{pi}{x}|_{1}^{infty}=pi.$$
is finite.
I extend it so its radius will be:
$$y=frac{1}{sqrt{x^2+c}},c is a constant.$$
Its surface will be:
$$int_{1}^{infty}frac{2pi}{sqrt{x^2+c}}dx=2piln left|frac{x}{sqrt{c}}+sqrt{frac{c+x^2}{c}}right||_1^inftyrightarrow diverges.$$
Then its volume:
$$int_{1}^{infty}frac{pi}{x^2+c}dx=frac{pi ^2}{2sqrt{c}}-frac{pi arctan left(sqrt{frac{1}{c}}right)}{sqrt{c}}rightarrow converges.$$
I tried to extend further for the radius:
$$y=frac{1}{sqrt[3]{x^3+c}}.\$$
Its surface:
$$int_{1}^{infty} frac{2pi}{sqrt[3]{x^3+c}} dx.\$$
Its volume:
$$ int_{1}^{infty} frac{pi}{(x^3+c)^{2/3}} dx.\$$
I can't find a general formula for
$$int_ frac{1}{sqrt[3]{x^3+c}} dx.\$$
If c is polynomial and its degree will be less then the x term before it.
How would I do a general expression for:
$$int_ frac{1}{sqrt[n]{ax^n+bx^{n-1}+cdots+z}} dx.\$$
polynomials soft-question improper-integrals indefinite-integrals
edited yesterday
Harry Peter
5,45111439
asked Dec 24 at 6:58
John He
94
I find that the Torricelli's trumpet has very interesting properties.
It's surface:
$$int_{1}^{infty}frac{2pi}{x}dx=2piln{x}|_{1}^{infty}rightarrowinfty.$$
is infinity.
Its volume:
$$int_{1}^{infty}frac{pi}{x^{2}}dx=-frac{pi}{x}|_{1}^{infty}=pi.$$
is finite.
I extend it so its radius will be:
$$y=frac{1}{sqrt{x^2+c}},c is a constant.$$
Its surface will be:
$$int_{1}^{infty}frac{2pi}{sqrt{x^2+c}}dx=2piln left|frac{x}{sqrt{c}}+sqrt{frac{c+x^2}{c}}right||_1^inftyrightarrow diverges.$$
Then its volume:
$$int_{1}^{infty}frac{pi}{x^2+c}dx=frac{pi ^2}{2sqrt{c}}-frac{pi arctan left(sqrt{frac{1}{c}}right)}{sqrt{c}}rightarrow converges.$$
I tried to extend further for the radius:
$$y=frac{1}{sqrt[3]{x^3+c}}.\$$
Its surface:
$$int_{1}^{infty} frac{2pi}{sqrt[3]{x^3+c}} dx.\$$
Its volume:
$$ int_{1}^{infty} frac{pi}{(x^3+c)^{2/3}} dx.\$$
I can't find a general formula for
$$int_ frac{1}{sqrt[3]{x^3+c}} dx.\$$
If c is polynomial and its degree will be less then the x term before it.
How would I do a general expression for:
$$int_ frac{1}{sqrt[n]{ax^n+bx^{n-1}+cdots+z}} dx.\$$
polynomials soft-question improper-integrals indefinite-integrals
polynomials soft-question improper-integrals indefinite-integrals
edited yesterday
Harry Peter
5,45111439
asked Dec 24 at 6:58
John He
94
edited yesterday
Harry Peter
5,45111439
asked Dec 24 at 6:58
John He
94
edited yesterday
Harry Peter
5,45111439
edited yesterday
Harry Peter
5,45111439
edited yesterday
Harry Peter
5,45111439
5,45111439
asked Dec 24 at 6:58
John He
94
asked Dec 24 at 6:58
John He
94
asked Dec 24 at 6:58
John He
94
94
1
It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28
add a comment |
1
It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28
1
1
It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28
add a comment |
1 Answer
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General indefinite integrals of the form
$$int frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}, qquad a_n neq 0,$$
do not have closed forms when $n > 2$.
For the particular integral $$int frac{dx}{sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = frac{c}{1 - u^3}$$ gives
$$int frac{dx}{sqrt[3]{x^3 + c}} = int frac{u ,du}{1 - u^3} ,$$
which can be handled by standard techniques.
On the other hand, as $x to infty$, $frac{1}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$frac{1}{sqrt[n]{a_n}} frac{1}{x},$$ and so the definite integrals of the form $$int_M^{infty} frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.
answered 2 days ago
Travis
59.5k767146
add a comment |
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General indefinite integrals of the form
$$int frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}, qquad a_n neq 0,$$
do not have closed forms when $n > 2$.
For the particular integral $$int frac{dx}{sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = frac{c}{1 - u^3}$$ gives
$$int frac{dx}{sqrt[3]{x^3 + c}} = int frac{u ,du}{1 - u^3} ,$$
which can be handled by standard techniques.
On the other hand, as $x to infty$, $frac{1}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$frac{1}{sqrt[n]{a_n}} frac{1}{x},$$ and so the definite integrals of the form $$int_M^{infty} frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.
answered 2 days ago
Travis
59.5k767146
add a comment |
General indefinite integrals of the form
$$int frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}, qquad a_n neq 0,$$
do not have closed forms when $n > 2$.
For the particular integral $$int frac{dx}{sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = frac{c}{1 - u^3}$$ gives
$$int frac{dx}{sqrt[3]{x^3 + c}} = int frac{u ,du}{1 - u^3} ,$$
which can be handled by standard techniques.
On the other hand, as $x to infty$, $frac{1}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$frac{1}{sqrt[n]{a_n}} frac{1}{x},$$ and so the definite integrals of the form $$int_M^{infty} frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.
answered 2 days ago
Travis
59.5k767146
add a comment |
General indefinite integrals of the form
$$int frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}, qquad a_n neq 0,$$
do not have closed forms when $n > 2$.
For the particular integral $$int frac{dx}{sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = frac{c}{1 - u^3}$$ gives
$$int frac{dx}{sqrt[3]{x^3 + c}} = int frac{u ,du}{1 - u^3} ,$$
which can be handled by standard techniques.
On the other hand, as $x to infty$, $frac{1}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$frac{1}{sqrt[n]{a_n}} frac{1}{x},$$ and so the definite integrals of the form $$int_M^{infty} frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.
answered 2 days ago
Travis
59.5k767146
General indefinite integrals of the form
$$int frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}, qquad a_n neq 0,$$
do not have closed forms when $n > 2$.
For the particular integral $$int frac{dx}{sqrt[3]{x^3 + c}} ,$$ however, substituting $$x^3 = frac{c}{1 - u^3}$$ gives
$$int frac{dx}{sqrt[3]{x^3 + c}} = int frac{u ,du}{1 - u^3} ,$$
which can be handled by standard techniques.
On the other hand, as $x to infty$, $frac{1}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$ is comparable (for $n > 0$) to $$frac{1}{sqrt[n]{a_n}} frac{1}{x},$$ and so the definite integrals of the form $$int_M^{infty} frac{dx}{sqrt[n]{a_n x^n + cdots + a_1 x + a_0}}$$ arising in the surface area computations all diverge.
answered 2 days ago
Travis
59.5k767146
answered 2 days ago
Travis
59.5k767146
answered 2 days ago
Travis
59.5k767146
answered 2 days ago
Travis
59.5k767146
59.5k767146
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It's not an elementary function, you'll need elliptic integrals for that.
– Hans Lundmark
Dec 24 at 8:46
@John He - It seems this is your first post. On MSE, it is required to provide context and background to your question and to speak to whatever methods you have employed (and provide the detail of that). If you haven't tried anything, you must be upfront and ask for a starting point. Make sure you tag the 'Soft Question' tag when you do so. It seems over the top when you first start on the site, but over time you come to realise that it's a great system. Please message me if you have any questions!
– DavidG
Dec 26 at 4:28