Definite integration

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girishpimoli

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girishpimoli

#1 h Yesterday at 4:21 AM • 2 Y

Y by MihaiT, LawofCosine

Evaluation of $\displaystyle \int^1_0 \frac{x^2+3}{x^4+10x^2+5}dx$

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LawofCosine

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LawofCosine

#2 h Yesterday at 6:58 AM

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maybe complete the square on the denominator and try trig substitution? I not sure though....

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Amkan2022

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Amkan2022

#3 h Yesterday at 7:07 AM

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PFD then $u = \frac{x}{\sqrt{2\sqrt{5}+5}}$ Sub works. We get something in $\arctan(u)$ , but the result seems pretty ugly (radicals)

This post has been edited 1 time. Last edited by Amkan2022, Yesterday at 7:07 AM

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LawofCosine

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LawofCosine

#4 h Yesterday at 7:52 AM

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what is PFD?

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awzhang10

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awzhang10

#5 h Yesterday at 3:15 PM

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partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

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vanstraelen

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vanstraelen

#6 h Yesterday at 4:52 PM

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$I=\int \frac{x^2+3}{x^4+10x^2+5}\ dx=\sqrt{\frac{5-\sqrt{5}}{50}}\arctan \sqrt{1-\frac{2}{\sqrt{5}}} \cdot x +\sqrt{\frac{5+\sqrt{5}}{50}}\arctan \sqrt{1+\frac{2}{\sqrt{5}}} \cdot x + C$

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HockeyMaster85

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HockeyMaster85

#7 h Yesterday at 10:39 PM

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awzhang10 wrote:

partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

are you sure? what about like $\int \frac{dx}{x^5 + x + 1}$ ?

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aidan0626

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aidan0626

#8 h Yesterday at 10:55 PM

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HockeyMaster85 wrote:

awzhang10 wrote:

partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

are you sure? what about like $\int \frac{dx}{x^5 + x + 1}$ ?

wolframalpha gives a closed form for that

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LawofCosine

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LawofCosine

#9 h Yesterday at 11:12 PM

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awzhang10 wrote:

partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

I think you mean partial fraction decomposition, but thanks!

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HacheB2031

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HacheB2031

#10 h Today at 2:01 AM

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HockeyMaster85 wrote:

awzhang10 wrote:

partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

are you sure? what about like $\int \frac{dx}{x^5 + x + 1}$ ?

$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$ (courtesy of Wolfram Alpha). I believe you mean the polynomial $x^5-x+1,$ as it doesn't have any roots that can be represented exactly and therefore cannot be factored exactly without having one of the roots defined as, well, its root. AND STILL, Wolfram gives the closed-form $\int\frac{\text dx}{x^5-x+1}=\sum_{\{\omega:\omega^5-\omega+1=0\}}\left[\frac{\log(x-\omega)}{5\omega^4-1}\right]+C,$ where $\{\omega:\omega^5-\omega+1=0\}$ is the set of all roots of $\omega^5-\omega+1=0$ and the sum cycles through all $5$ and gives a term for each. So it is even possible to integrate a rational function with closed-form even when it has poles that are unable to be located exactly.

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HockeyMaster85

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HockeyMaster85

#11 h Today at 3:34 AM

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HacheB2031 wrote:

HockeyMaster85 wrote:

awzhang10 wrote:

partial fraction denominator
there is a fairly famous result if i remember correctly that says any rational function can be integrated in a closed form

are you sure? what about like $\int \frac{dx}{x^5 + x + 1}$ ?

$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$ (courtesy of Wolfram Alpha). I believe you mean the polynomial $x^5-x+1,$ as it doesn't have any roots that can be represented exactly and therefore cannot be factored exactly without having one of the roots defined as, well, its root. AND STILL, Wolfram gives the closed-form $\int\frac{\text dx}{x^5-x+1}=\sum_{\{\omega:\omega^5-\omega+1=0\}}\left[\frac{\log(x-\omega)}{5\omega^4-1}\right]+C,$ where $\{\omega:\omega^5-\omega+1=0\}$ is the set of all roots of $\omega^5-\omega+1=0$ and the sum cycles through all $5$ and gives a term for each. So it is even possible to integrate a rational function with closed-form even when it has poles that are unable to be located exactly.

yeah thats what i meant. u count that as a closed form? its a summation which isnt a closed form, and with roots that are nasty

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aidan0626

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aidan0626

#12 h Today at 3:37 AM

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HockeyMaster85 wrote:

yeah thats what i meant. u count that as a closed form? its a summation which isnt a closed form, and with roots that are nasty

it is a finite sum

and just bc it's nasty doesn't mean it's not a closed form

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rchokler

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rchokler

#13 h Today at 4:12 AM

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$x^4+10x+5=(x^2+5)^2-20=\left(x^2+5+2\sqrt{5}\right)\left(x^2+5-2\sqrt{5}\right)$

$\frac{x^2+3}{x^4+10x^2+5}=\frac{Ax+B}{x^2+5+2\sqrt{5}}+\frac{Cx+D}{x^2+5-2\sqrt{5}}=\frac{(A+C)x^3+(B+D)x^2+[(5-2\sqrt{5})A+(5+2\sqrt{5})C]x+[(5-2\sqrt{5})B+(5+2\sqrt{5})D]}{x^4+10x+5}$

$\begin{cases}A+C=0\\B+D=1\\(5-2\sqrt{5})A+(5+2\sqrt{5})C=0\\(5-2\sqrt{5})B+(5+2\sqrt{5})D=3\end{cases}\implies\begin{cases}C=-A\\D=1-B\\(-4\sqrt{5})A=0\\(-4\sqrt{5})B=-2-2\sqrt{5}\end{cases}\implies\begin{cases}A=0\\B=\frac{5+\sqrt{5}}{10}\\C=0\\D=\frac{5-\sqrt{5}}{10}\end{cases}$

$\int_0^1\frac{x^2+3}{x^4+10x^2+5}\ dx=\frac{5+\sqrt{5}}{10}\int_0^1\frac{dx}{x^2+5+2\sqrt{5}}+\frac{5-\sqrt{5}}{10}\int_0^1\frac{dx}{x^2+5-2\sqrt{5}}=\left.\frac{5+\sqrt{5}}{10\sqrt{5+2\sqrt{5}}}\arctan\frac{x}{\sqrt{5+2\sqrt{5}}}+\frac{5-\sqrt{5}}{10\sqrt{5-2\sqrt{5}}}\arctan\frac{x}{\sqrt{5-2\sqrt{5}}}\right|_0^1$ $=\frac{5+\sqrt{5}}{10\sqrt{5+2\sqrt{5}}}\arctan\frac{1}{\sqrt{5+2\sqrt{5}}}+\frac{5-\sqrt{5}}{10\sqrt{5-2\sqrt{5}}}\arctan\frac{1}{\sqrt{5-2\sqrt{5}}}$

This post has been edited 2 times. Last edited by rchokler, Today at 4:13 AM

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Ninjametry

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Ninjametry

#14 h 2 hours ago

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A Much simpler solution will be when
x-3/x= t
divide the numerator and denominator by x^2
now express in t and solve

This post has been edited 1 time. Last edited by Ninjametry, 2 hours ago

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