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Is there any way to evaluate: $$\int\frac{t}{\sqrt{t^{4}-1}}\text dt$$

I have started by using: $y=\sqrt{t^4-1}$

then: $x=\sqrt{y^{2}+1}$

and finally: $x = \cosh(t)$

but I didn't get any result, any tips?

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    Did you consider using a halfway step of $x=t^2,\text dx = 2t\text dt$? – 2017-02-26

2 Answers 2

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Well, you have: $$\int \frac{t}{\sqrt{t^4-1}}~dt$$ You can substitute $u=t^2$ and $du=2t~dt$ to obtain: $$\frac{1}{2}\int \frac{1}{\sqrt{u^2-1}}~du$$ Which is a well known integral. If you want to derive the solution to this well known integral, consider using the substitution $u=\sec v$ and $du=\sec v\cdot \tan v~dv$.

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Let $t^2=\cosh x$ and the integral becomes $$\frac 12\int 1 dx=\frac 12\operatorname{arcosh}(t^2)+c$$

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    I can't vote twice, but +1 for skipping my "halfway step" altogether... – 2017-02-26
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    if $t^2=\cosh x$ then $dt =?$ I mean in general if you have : $g(t)=f(x)$ then how to evaluate $dt$ ? – 2017-02-26
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    $2tdt=\sinh xdx$ so you replace $tdt$ together... – 2017-02-26
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    ok thank you , so in general we have $g'(t)dt=f'(x)dx$ isn'it ? – 2017-02-26
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    Yes that's it.. – 2017-02-26