Definite integration

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Definite integration

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#1 h Monday 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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#2 h Monday 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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#3 h Monday 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, Monday at 7:07 AM

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#4 h Monday at 7:52 AM

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what is PFD?

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#5 h Monday 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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#6 h Monday 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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#7 h Monday 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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#8 h Monday 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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#9 h Monday 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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#10 h Yesterday 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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#11 h Yesterday 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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#12 h Yesterday 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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#13 h Yesterday 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, Yesterday at 4:13 AM

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#14 h Yesterday at 12:43 PM

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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, Yesterday at 12:43 PM

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#15 h Yesterday at 8:50 PM

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@Ninjametry

Please, show the full solution using your substitution.

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