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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.

cdummie
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3 Answers3

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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.

egreg
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4

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


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

Jan Eerland
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  • Thanks a lot, anyway, i am wondering, are there any non-trigonometric substitutions that can be used to solve this integral? – cdummie Mar 02 '16 at 13:46
  • @cdummie You're welcome. This is the most easy way, of solving this integral! – Jan Eerland Mar 02 '16 at 13:49
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    Always easy with mathematica, isn't it? :) – Laplacian Mar 02 '16 at 13:51
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    @1over137 any reason here to suspect this is a mathematica based answer? – tired Mar 02 '16 at 14:03
  • @tired Yes: it's identical to the way in which Mathematica acts for solving these kinds of integrals (and also, after suspecting, I checked and the passages are literally identical step by step). I'm not against using Mathematica, I'm against those who pretend they solved integrals whilst they just copied.. Sigh! See the other user's answer, it's awesome! – Laplacian Mar 02 '16 at 14:05
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    @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! – Jan Eerland Mar 02 '16 at 14:07
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    @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. – tired Mar 02 '16 at 14:15
  • @tired exactly! – Jan Eerland Mar 02 '16 at 14:26
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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.

Julian Rosen
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