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Frankly, i don't have a solution to this, not even incorrect one, but, this integral looks a lot like that standard type of integral $I=\int\frac{Mx+N}{(x-\alpha)^n\sqrt{ax^2+bx+c}}$ which can be solved using substitution $x-\alpha=\frac{1}{t}$ so i tried to find such subtitution that will make this integral completely the same as this standard integral so i could use substitution i mentioned, so i tried two following substitutions

$x^2-4=t^2 \Rightarrow x^2=t^2+4 \Rightarrow x=\sqrt{t^2+4}$ then i had to determine $dx$

$2xdx=2tdt \Rightarrow dx=\frac{tdt}{\sqrt{t^2+4}}$

from here i got:

$\int\frac{dt}{(t^2+5)\sqrt{(t^2+4)}}$ but i have no idea what could i do with this, so i tried different substitution

$x^2+1=t^2$ and then, by implementing the same pattern i used with the previous substitution i got this integral

$\int\frac{dt}{t\sqrt{(t^2-1)(t^2-5)}}$ but again, i don't know what to do with this, so i could use some help.

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    $\begingroup$ Non-trigonometric substitution (that often works in such cases): $t=x/\sqrt{x^2-4}$. $\endgroup$ – mickep Mar 2 '16 at 13:59
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    $\begingroup$ Interesting that such a standard integral gains this amount of upvotes ^^ $\endgroup$ – tired Mar 2 '16 at 14:02
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Use $x=2\cosh t$, so the integral becomes $$ \int\frac{1}{1+4\cosh^2t}\,dt= \int\frac{1}{1+e^{2t}+2+e^{-2t}}= \int\frac{e^{2t}}{e^{4t}+3e^{2t}+1}\,dt $$ and then set $u=e^{2t}$ so you get $$ \frac{1}{2}\int\frac{1}{u^2+3u+1}\,du $$ that can be computed by partial fractions.

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    $\begingroup$ By far the easiest way! +1 $\endgroup$ – zz20s Mar 2 '16 at 13:58
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    $\begingroup$ This is an awesome answer!! I like people like you!! ^^ $\endgroup$ – Von Neumann Mar 2 '16 at 14:05
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    $\begingroup$ @1over137 this can be as easily as the other answer a mathematica assisted proof (i don't believe that, but i hope u see the point) $\endgroup$ – tired Mar 2 '16 at 14:16
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    $\begingroup$ @tired No assistance except paper and pencil. $\endgroup$ – egreg Mar 2 '16 at 15:09
  • $\begingroup$ @egreg , i'm absolutely sure about that, my comment was related to a discussion concerning the other answer :) $\endgroup$ – tired Mar 2 '16 at 15:23
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HINT:

Your integral:

$$\text{I}=\int\frac{1}{\left(x^2+1\right)\sqrt{x^2-4}}\space\text{d}x=$$


Subsitute $x=2\sec(u)$ and $\text{d}x=2\tan(u)\sec(u)\space\text{d}u$.

Then $\sqrt{x^2-4}=\sqrt{4\sec^2(u)-4}=2\tan(u)$ and $u=\text{arcsec}\left(\frac{x}{2}\right)$:


$$\int\frac{\sec(u)}{4\sec^2(u)+1}\space\text{d}u=\int\frac{\cos(u)}{5-\sin^2(u)}\space\text{d}u=$$


Substitute $s=\sin(u)$ and $\text{d}s=\cos(u)\space\text{d}u$:


$$\int\frac{1}{5-s^2}\space\text{d}s=\frac{1}{5}\int\frac{1}{1-\frac{s^2}{5}}\space\text{d}s=$$


Substitute $p=\frac{s}{\sqrt{5}}$ and $\text{d}p=\frac{1}{\sqrt{5}}\space\text{d}s$:


$$\frac{1}{\sqrt{5}}\int\frac{1}{1-p^2}\space\text{d}p=\frac{\text{arctanh}\left(p\right)}{\sqrt{5}}+\text{C}=$$ $$\frac{\text{arctanh}\left(\frac{s}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sin\left(u\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=$$ $$\frac{\text{arctanh}\left(\frac{\sin\left(\text{arcsec}\left(\frac{x}{2}\right)\right)}{\sqrt{5}}\right)}{\sqrt{5}}+\text{C}=\frac{\text{arctanh}\left(\frac{\sqrt{x^2-4}}{x\sqrt{5}}\right)}{\sqrt{5}}+\text{C}$$

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  • $\begingroup$ Thanks a lot, anyway, i am wondering, are there any non-trigonometric substitutions that can be used to solve this integral? $\endgroup$ – cdummie Mar 2 '16 at 13:46
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    $\begingroup$ Always easy with mathematica, isn't it? :) $\endgroup$ – Von Neumann Mar 2 '16 at 13:51
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    $\begingroup$ @1over137 any reason here to suspect this is a mathematica based answer? $\endgroup$ – tired Mar 2 '16 at 14:03
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    $\begingroup$ @1over137 You can think that, that is the case, but you are wrong, because I didn't use Mathematica for this question, but you can think that, it's totally fine by me! So as tired tried to say, maybe you've to say; "I think he used Matematica" because this is not the way you talk about others, I'm sorry! $\endgroup$ – Jan Mar 2 '16 at 14:07
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    $\begingroup$ @1over137 the way i use mathematica produces intermediate steps which are slightly different. So if i would have posted this as an answer u would never had the idea that i'm "cheating"... i think that shows that ur accusations are a little bit off.furhtermore, if Jan says he didn't use mathematica we should believe him unless we have a definite proof that this is the case. $\endgroup$ – tired Mar 2 '16 at 14:15
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The curve $t^2=x^2-4$ is a hyperbola, which can be parametrized by a single value. Rewrite this equation as $(x+t)(x-t)=4$, and set $y=x+t$. Then $4/y=x-t$, and we have $$ x=\frac{y+\frac{4}{y}}{2},\;\;\;\;t=\frac{y-\frac{4}{y}}{2}. $$ Compute $dx=(1/2-2 y^{-2})dy$, and the integral becomes $$ \int \frac{dx}{(x^2+1)\sqrt{x^2-4}}=\int \frac{\frac{1}{2}-\frac{2}{y^2}}{\left(\left(\frac{y+\frac{4}{y}}{2}\right)^2+1\right)\left(\frac{y-\frac{4}{y}}{2}\right)}dy=\int \frac{4y\, dy}{y^4+12 y^2+16}. $$ This can be computed using partial fractions.

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