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I am trying to prove that

$$\displaystyle \int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$$

$u=(x^2+1)^{1/2}$ then $du=(x^2+1)^{-1/2}dx$

$$\int_{1}^{\infty}{u^2+(u^2-1)\sqrt{u^2+1}\over u^2\sqrt{u^2+1}}du$$

$v=(u^2+1)^{1/2}$ then $dv=(u^2+1)^{-1/2}du$

$$\int_{1}^{\infty}{v^3+v^2-2v-1\over v^2-1}dv$$

$\int_{1}^{\infty}{v^2\over v^2-1}-{1\over v^2-1}dv$ -$\ln{(v^2-1)}|_{1}^{\infty}$

I am sure I when wrong somewhere, but I can figured it out.

Any help?

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  • $\begingroup$ du and dv are not correct - you are missing a multiplication by x and u respectively. $\endgroup$ – Paul Jan 9 '17 at 12:57
  • $\begingroup$ Your first differentiation is wrong: $$u=\sqrt{x^2+1}\implies du=\frac x{\sqrt{x^2+1}}dx$$ $\endgroup$ – DonAntonio Jan 9 '17 at 12:57
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    $\begingroup$ In other question you also had a mistake differentiating. You should really take good care of this as it is a rather bad idea, imo, to try to tackle antiderivatives and integration without mastering differentiation. $\endgroup$ – DonAntonio Jan 9 '17 at 13:00
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We have $$I=\int_{0}^{\infty}\frac{\sqrt{x^{2}+1}+x^{2}\sqrt{x^{2}+2}}{\sqrt{\left(x^{2}+1\right)\left(x^{2}+2\right)}}\frac{1}{\left(x^{2}+1\right)^{2}}dx=\int_{0}^{\infty}\frac{x^{2}}{\left(x^{2}+1\right)^{5/2}}dx+\int_{0}^{\infty}\frac{1}{\left(x^{2}+1\right)^{2}\sqrt{x^{2}+2}}dx=I_{1}+I_{2}. $$ Now recalling that the Beta function has the representation $$\int_{0}^{\infty}\frac{u^{m}}{\left(u+1\right)^{m+n+2}}du=B\left(m+1,n+1\right) $$ we get $$I_{1}=\frac{1}{2}\int_{0}^{\infty}\frac{u^{1/2}}{\left(u+1\right)^{5/2}}du=\frac{B\left(\frac{3}{2},1\right)}{2}=\frac{1}{3}. $$ If you prefer, you can also compute $I_{1} $ using the substitution $x=\tan\left(v\right) $. For $I_{2} $ we get $$I_{2}\stackrel{x=\sqrt{2}\tan\left(v\right)}{=}\int_{0}^{\pi/2}\frac{\left(1-\sin^{2}\left(v\right)\right)\cos\left(v\right)}{\left(\sin^{2}\left(v\right)+1\right)^{2}}dv $$ $$\stackrel{z=\sin\left(v\right)}{=}\int_{0}^{1}\frac{1-z^{2}}{\left(z^{2}+1\right)^{2}}dz=\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz-\int_{0}^{1}\frac{1}{z^{2}+1}dz $$ and note that $$\int_{0}^{1}\frac{2}{\left(z^{2}+1\right)^{2}}dz\stackrel{z=\tan\left(w\right)}{=}2\int_{0}^{\pi/4}\cos^{2}\left(w\right)dw=\frac{1}{2}+\frac{\pi}{4} $$ hence $$I=\frac{1}{3}+\frac{\pi}{4}+\frac{1}{2}-\frac{\pi}{4}=\color{red}{\frac{5}{6}}.$$

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    $\begingroup$ (+1) It is good to be familiar with the Beta function! In this case, a primitive of the first part is given by $(1/3)x^3/(1+x^2)^{3/2}$ and $x\sqrt{2+x^2}/(2(1+x^2))$ for the second part. $\endgroup$ – mickep Jan 9 '17 at 14:46
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    $\begingroup$ Q.E.D. (+1). People like you @Marco Cantarini do inspired me to study maths one day. $\endgroup$ – user339807 Jan 9 '17 at 14:55
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Let $A=\sqrt{x^2+1}$ and $B=\sqrt{x^2+2}$ so that have $A^2-1=x^2$. With these you have

\begin{align} {\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}&=\frac{A+(A^2-1)B}{AB}\frac{1}{A^4}\\ &=\frac{1}{BA^4}-\frac{1}{A^3}-\frac{1}{A^5}\\ \end{align}

Note that the integral of all of these terms is very simple to evaluate, so have fun :-)

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on primitive function should be $${\frac {\sqrt {{x}^{2}+1}\sqrt {{x}^{2}+2}}{\sqrt { \left( {x}^{2}+1 \right) \left( {x}^{2}+2 \right) }}\arctan \left( {\frac {x}{\sqrt { {x}^{2}+2}}} \right) }+{\frac {\sqrt {{x}^{2}+2} \left( {\rm arcsinh} \left(x\right)\sqrt {{x}^{2}+1}-x \right) }{\sqrt { \left( {x}^{2}+1 \right) \left( {x}^{2}+2 \right) }}} $$

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  • $\begingroup$ According to Mathematica, a primitive is given by $$\frac{x\sqrt{x^2+2}}{2 \left(x^2+1\right)}+\frac{x^3}{3 \left(x^2+1\right)^{3/2}}.$$ But isn't it the way to get there that is interesting? $\endgroup$ – mickep Jan 9 '17 at 13:38
  • $\begingroup$ yea we can compute both the first derivative, is this ok? $\endgroup$ – Dr. Sonnhard Graubner Jan 9 '17 at 13:39
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    $\begingroup$ I dare you, please don't forget FullSimplify[] at the end of your mathematica evaluation.. $\endgroup$ – tired Jan 9 '17 at 15:51
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    $\begingroup$ @tired, I think the Dr uses maple, in fact. $\endgroup$ – mickep Jan 9 '17 at 21:19
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$u=(x^2+1)^{\frac{1}{2}}$, thus $du = \frac{1}{2}(x^2+1)^{-\frac{1}{2}}2xdx =\frac{x}{(x^2+1)^{\frac{1}{2}}}dx$.

$x \in (0, +\infty), u \in (1, +\infty), x =(u^2-1)^{\frac{1}{2}}$

$dx = \frac{(x^2+1)^{1/2}}{x}du=\frac{u}{(u^2-1)^{1/2}}du$

$\int_0^\infty\frac{u+(u^2-1)(u^2+1)^{1/2}}{u^2(u^2+1)^{1/2}} \cdot \frac{1}{u^4} \cdot \frac{u}{(u^2-1)^{1/2}}du$

I find some mistakes, but I can not solve it. Sorry for that.

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