I don"t know how to solve the very first right hand side integral. I tried to let $u^2 - 1 = t$, so it becomes

$$intfrac1sqrtu^2 - 1=intfrac1sqrttfrac12u , dt = int frac1sqrtt frac12 frac1sqrtt-1 , dt,$$

which doesn"t seem to work. The correct answer appears to it is in

$$ln left(sqrtx^2 - 1 - x ight)$$

But just how do we get this?




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edited Sep 8 "16 in ~ 23:28
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$$int frac dx sqrt x^ 2 -1 =int frac sinh t sinh t dt=t +C\ \ x=cosh t \ dx=sinh t \ x=frac e ^ t + e ^ -t 2 \ 2x e ^ t = e ^ 2t +1\ e ^ 2t -2x e ^ t +1=0\ e ^ t =x+sqrt x ^ 2 -1 $$

$$t=ln $$


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answer Sep 8 "16 in ~ 22:27
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haqnaturalhaqnatural
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LEt $u=cosh t$. You may need to uncover a formula because that arccosh x


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answer Sep 8 "16 at 22:20
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Let $x=sec t, ; dx=sec t an t,dt$ come get

$displaystyleintfrac1sqrtx^2-1dx=intfracsec t an t an tdt=intsec t, dt=lnig|sec t+ an tig|+C=lnig|x+sqrtx^2-1ig|+C$


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answer Sep 8 "16 at 23:06
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If the hyperbolic trigonometry is not known, we have the right to see that

$intfracdtsqrttsqrtt+1=intfracsqrtt+1sqrtt-fracsqrt tsqrtt+1dt.$

Then $z=1+frac1t$ gives $t=frac1z-1$, $dt=frac-dz(1-z)^2$ and

$frac12intfracsqrtt+1sqrtt-fracsqrt tsqrtt+1dt=frac12intfrac(1-z)dzsqrtz(1-z)^2=frac12intfracdzsqrtz(1-z)=frac14intfrac1sqrt z(1-sqrtz)+frac1sqrt z(1+sqrtz)dz.$

It comes

$intfracdusqrtu^2-1=frac12left(ln(1+sqrt z)-ln(sqrt z-1) ight)=frac12lnleft(frac1+sqrt zsqrt z-1 ight)=frac12lnleft(fracsqrt t+sqrt1+tsqrt1+t-sqrt t ight)=frac12lnleft(fracu+sqrtu^2-1u-sqrtu^2-1 ight)$

and as $lnleft(frac1u-sqrtu^2-1 ight)=ln(u+sqrtu^2-1)$, we attain $intfracdusqrtu^2-1=ln(u+sqrtu^2-1)$.

We can likewise note the $intfracdusqrtu^2-1=frac12intfracsqrtu+1sqrtu-1-fracsqrtu-1sqrt1+udu$ and take the substitution $t=fracu+1u-1$.