Another Integral

**Amer** · February 8th 2011, 11:20 AM

$\displaystyle \int_{0}^{\infty} \frac{2x^2+3}{x^4+7x^2+1} dx$

any hints
I tried much I tried by parts considering dx=dv u=f(x) failed
I tried sub u=x^2 failed
parts dv=2x^2+3 u=1/(x^4+7x^2+1) failed and another things but I failed at all

Thanks

**dwsmith** · February 8th 2011, 11:39 AM

$\displaystyle\int_0^{\infty}\frac{2x^2}{x^4+7x^2+1 } \ dx$

$\displaystyle u=\tan^{-1}\left(\frac{(\sqrt{5}-3)x}{2}\right)$

$\displaystyle 3\int_0^{\infty}\frac{1}{x^4+7x^2+1} \ dx$

$\displaystyle u_2=\tan^{-1}\left(\frac{(\sqrt{5}+3)x}{2}\right)$

**Amer** · February 8th 2011, 11:52 AM

can I know how u choose this sub specially the constant multiply with x in the arctan
although it is still messy :S

**dwsmith** · February 8th 2011, 11:54 AM

Originally Posted by Amer

can I know how u choose this sub specially the constant multiply with x in the arctan
although it is still messy :S

I looked at the solutions to the separate indefinite integrals. There is probably an easier way to do it though.

**tonio** · February 8th 2011, 11:58 AM

Originally Posted by Amer

$\displaystyle \int_{0}^{\infty} \frac{2x^2+3}{x^4+7x^2+1} dx$

any hints
I tried much I tried by parts considering dx=dv u=f(x) failed
I tried sub u=x^2 failed
parts dv=2x^2+3 u=1/(x^4+7x^2+1) failed and another things but I failed at all

Thanks

Unless there's some snaky trick, this seems to be a specially nasty exercise in decomposition of rational functions:

$\displaystye{x^4+7x^2+1=\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}$ , so :

$\displaystyle{\frac{2x^2+3}{x^4+7x^2+1}=\frac{Ax+B }{x^2+\frac{7-3\sqrt{5}}{2}}+\frac{Cx+D}{x^2+\frac{7+3\sqrt{5}}{ 2}}$ , and now multiply

by the common denominator, choose nice values of x in both sides and find the coefficients

A,B,C,D. It looks horrible, indeed...but who knows? Perhaps there's another way...

Tonio

**Amer** · February 8th 2011, 12:01 PM

Thanks I will work on it

$\displaystyle{\frac{2x^2+3}{x^4+7x^2+1}=\frac{Ax+B }{x^2+\frac{7-3\sqrt{5}}{2}}+\frac{Cx+D}{x^2+\frac{7+3\sqrt{5}}{ 2}}$

I found that

$B = \frac{3\sqrt{5} - 4}{3\sqrt{5}}$

$D = \frac{3\sqrt{5} +4}{3\sqrt{5}}$

$\displaystyle \int_{0}^{\infty} \frac{\frac{3\sqrt{5} - 4}{3\sqrt{5}}}{x^2 + \frac{7-3\sqrt{5}}{2}} + \frac{\frac{3\sqrt{5} +4}{3\sqrt{5}}}{x^2 + \frac{7+3\sqrt{5}}{2}} dx$

$\displaystyle \int_{0}^{\infty} \frac{\frac{3\sqrt{5} - 4}{3\sqrt{5}}}{x^2 + \frac{7-3\sqrt{5}}{2}} dx + \int_{0}^{\infty} \frac{\frac{3\sqrt{5} +4}{3\sqrt{5}}}{x^2 + \frac{7+3\sqrt{5}}{2}} dx$

both are arctan thanks

**simplependulum** · February 8th 2011, 07:14 PM

Write

$\displaystyle 2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1)$ where $A,B$ are constants , solving it we obtain $\displaystyle A = -0.5 ~,~ B = 2.5$

We have

$\displaystyle \int \frac{2x^2 + 3}{x^4 + 7x^2 + 1}~dx$

$\displaystyle = -\frac{1}{2} \int \frac{x^2 - 1}{x^4 + 7x^2 + 1 }~dx + \frac{5}{2} \int \frac{x^2 + 1}{x^4 + 7x^2 + 1}~dx$

For the first integral , consider

$\displaystyle \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx$

$\displaystyle = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }$

Sub. $\displaystyle u = x+1/x$

$\displaystyle = \int \frac{du}{u^2+5} = \frac{1}{\sqrt{5}} \tan^{-1}(u/\sqrt{5})$

$\displaystyle = \frac{1}{\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right)$

For the second integral , you may find it is a similar trick

$\displaystyle \int \frac{1 + 1/x^2}{x^2 + 7 + 1/x^2 }~dx$

$\displaystyle = \int \frac{d(x-1/x)}{(x-1/x)^2 + 9}$

$\displaystyle = \frac{1}{3} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right)$

Therefore , the answer is

$\displaystyle - \frac{1}{2\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right) + \frac{5}{6} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right) + C$

**TheCoffeeMachine** · February 8th 2011, 07:19 PM

After Tonio's factorization, we can write the $2x^2+3$ in terms of the sum of the two quadratic factors, i.e:
We just observe that $\displaystyle \left(x^2+\frac{7-3\sqrt{5}}{2}\right)+\left(x^2+\frac{7+3\sqrt{5}}{ 2}\right) = 2x^2+7$ then subtract $4$ .

$\displaystyle \begin{aligned}I & = \int\frac{2x^2+3}{x^4+7x^2+1}\;{dx} \\ & = \int \frac{2x^2+3}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}\;{dx} \\ & = \int \frac{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)+\left(x^2+\frac{7+3\sqrt{5}}{ 2}\right)-4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}\;{dx} \\ & = \int \frac{1}{x^2+\frac{7+3\sqrt{5}}{2}}\;{dx}+ \int \frac{1}{x^2+\frac{7-3\sqrt{5}}{2}}\;{dx} -\int \frac{4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}\;{dx} \\ & = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )+\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)-I_{1}. \end{aligned}$

For $I_{1}$ , we write $4$ as in terms of the two quadratic factors, and we do it as follows -- let:

$\alpha\left(x^2+\frac{7+3\sqrt{5}}{2}\right)-\alpha\left(x^2+\frac{7-3\sqrt{5}}{2}\right)} = 4$ , then putting $x = 0$ gives you the value:

$\displaystyle \alpha = \frac{4}{\frac{1}{2}(7+3\sqrt{5})-\frac{1}{2}(7-3\sqrt{5})} = \frac{4}{\frac{1}{2}(3\sqrt{5}+3\sqrt{5})} = \frac{4}{3\sqrt{5}}.$ Therefore, we have:

$\displaystyle \begin{aligned} I_{1} & = \int \frac{4}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}\;{dx} \\ & = \int \frac{\frac{4}{3\sqrt{5}}\left(x^2+\frac{7+3\sqrt{ 5}}{2}\right)-\frac{4}{3\sqrt{5}}\left(x^2+\frac{7-3\sqrt{5}}{2}\right)}{\left(x^2+\frac{7-3\sqrt{5}}{2}\right)\left(x^2+\frac{7+3\sqrt{5}}{2 }\right)}{dx} \\ & = \frac{4}{3\sqrt{5}}\int\frac{1}{x^2+\frac{7-3\sqrt{5}}{2}} \;{dx} - \frac{4}{3\sqrt{5}}\int\frac{1}{x^2+\frac{7+3\sqrt {5}}{2}} \;{dx} \\ & = \frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)-\frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7+3\sqrt{5}}}\tan^ {-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )}. \end{aligned}$

Therefore, we have:

$\begin{aligned} I & = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )+\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right) \\& -\frac{4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)+\frac{4\sqrt{2}}{3\sqrt{5}\sqrt {7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right ) +k \\& = \frac{\sqrt{2}}{\sqrt{7+3\sqrt{5}}}\left(1+\frac{4 }{3\sqrt{5}}\right)\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right ) +\frac{\sqrt{2}}{\sqrt{7-3\sqrt{5}}}\left(1-\frac{4}{3\sqrt{5}}\right)\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right) \\& = \boxed{\frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5 }\sqrt{7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)+k} \end{aligned}$

Isn't that just an absolute beauty? Beats all the sunsets one can see or imagine, haha! Now:

$\displaystyle \begin{aligned} \int_0^{\infty}\frac{2x^2+3}{x^4+7x^2+1} \; dx & = \frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5}\sqrt{ 7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)\bigg|_{0}^{\infty} \\& = \lim_{x \to \infty}\bigg[\frac{3\sqrt{2}\sqrt{5}+4\sqrt{2}}{3\sqrt{5}\sqrt{ 7+3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7+3\sqrt{5}}}\right )+\frac{3\sqrt{2}\sqrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\tan^{-1}\left(\frac{\sqrt{2}x}{\sqrt{7-3\sqrt{5}}}\right)\bigg] \\& = \frac{\pi}{2}\left(\frac{3\sqrt{2}\sqrt{5}+4\sqrt{ 2}}{3\sqrt{5}\sqrt{7+3\sqrt{5}}}+\frac{3\sqrt{2}\s qrt{5}-4\sqrt{2}}{3\sqrt{5}\sqrt{7-3\sqrt{5}}}\right) = \frac{\pi}{2}\left(\frac{5}{3}\right) = \frac{5\pi}{6}. \end{aligned}$

**TheCoffeeMachine** · February 8th 2011, 08:30 PM

What's happening with my latex?

Originally Posted by simplependulum

$\displaystyle 2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1)$ where $A,B$ are constants , solving it we obtain $\displaystyle A = -0.5 ~,~ B = 2.5$

I'm shocked by how you tweaked the integral just to make it admit those couple of substitutions! Nice!

**Amer** · February 9th 2011, 10:10 AM

Originally Posted by simplependulum

Write

$\displaystyle 2x^2 + 3 = A(x^2 - 1) + B(x^2 + 1)$ where $A,B$ are constants , solving it we obtain $\displaystyle A = -0.5 ~,~ B = 2.5$

We have

$\displaystyle \int \frac{2x^2 + 3}{x^4 + 7x^2 + 1}~dx$

$\displaystyle = -\frac{1}{2} \int \frac{x^2 - 1}{x^4 + 7x^2 + 1 }~dx + \frac{5}{2} \int \frac{x^2 + 1}{x^4 + 7x^2 + 1}~dx$

For the first integral , consider

$\displaystyle \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx$

$\displaystyle = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }$

Sub. $\displaystyle u = x+1/x$

$\displaystyle = \int \frac{du}{u^2+5} = \frac{1}{\sqrt{5}} \tan^{-1}(u/\sqrt{5})$

$\displaystyle = \frac{1}{\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right)$

For the second integral , you may find it is a similar trick

$\displaystyle \int \frac{1 + 1/x^2}{x^2 + 7 + 1/x^2 }~dx$

$\displaystyle = \int \frac{d(x-1/x)}{(x-1/x)^2 + 9}$

$\displaystyle = \frac{1}{3} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right)$

Therefore , the answer is

$\displaystyle - \frac{1}{2\sqrt{5}} \tan^{-1}\left( \frac{x^2+1}{\sqrt{5} x } \right) + \frac{5}{6} \tan^{-1}\left( \frac{x^2 - 1}{3x} \right) + C$

wow thats cooool thanks for u all
I like the way u think thats very cool I learned much from u

**AllanCuz** · February 9th 2011, 12:26 PM

Originally Posted by simplependulum

Write

For the first integral , consider

$\displaystyle \int \frac{1 - 1/x^2}{x^2 + 7 + 1/x^2 }~dx$

$\displaystyle = \int \frac{d(x+1/x)}{(x+1/x)^2 + 5 }$

Edit- Oh wow. My bad. Too much to drink. Gotcha

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