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$$ \int \frac{\text{d}t}{t^2\sqrt{t-2} } $$

I know it can be calculated using somewhat complicated substitutions, but is there possibly some clever way of solving that type of integral? I don't even expect full solution, just ideas.

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  • $\begingroup$ Unlikely as it involves the arctan function. $\endgroup$ – amcalde Feb 12 '16 at 17:45
  • $\begingroup$ The substitutions aren't that hard.... Completing the square should work here after bringing the $t^2$ inside, a pretty basic calculus 1 problem $\endgroup$ – Brevan Ellefsen Feb 12 '16 at 17:51
  • $\begingroup$ Three answer appeared here without anyone suggesting a rationalizing substitution. I posted that below. $\endgroup$ – Michael Hardy Feb 12 '16 at 18:39
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Substitute $t=2\tan^2(u)+2$: $$ \begin{align} \int\frac{\mathrm{d}t}{t^2\sqrt{t-2}} &=\frac1{\sqrt2}\int\cos^2(u)\,\mathrm{d}u\\ &=\frac1{2\sqrt2}\int(1+\cos(2u))\,\mathrm{d}u\\ &=\frac1{2\sqrt2}\left(u+\frac12\sin(2u)\right)+C\\ &=\frac1{2\sqrt2}\left(u+\frac{\tan(u)}{1+\tan^2(u)}\right)+C\\ &=\frac1{2\sqrt2}\arctan\left(\sqrt{\frac{t-2}2}\right)+\frac12\frac{\sqrt{t-2}}t+C\\ \end{align} $$

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HInts

1. use $x = t-2$, $\text{d}x = \text{d}t$

2. You're now with $\int \frac{\text{d}x}{(x+2)^2\sqrt{x}}$

3. Now use $y = \sqrt{x}$, $\text{d}y = \frac{1}{2\sqrt{x}}\ \text{d}x$

4. You now get $2 \int \frac{1}{(y^2 + 2)^2}\ \text{d}y$

5. Use now $y = \sqrt{2} \tan(z)$, $\text{d}y = \sqrt{2}\sec^2(z)\ \text{d}z$ and arranging...

6. ...you obtain $\frac{1}{\sqrt{2}}\int \cos^2(z)\ \text{d}z$ which is trivial

7. Final Result:

$$\int \frac{\text{d} t}{t^2\sqrt{t-2}} = \frac{2\sqrt{t-2} + \sqrt{2}t \arctan\left(\sqrt{\frac{t-2}{2}}\right)}{4t}$$

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    $\begingroup$ you solved. I liked it. $\endgroup$ – L.F. Cavenaghi Feb 12 '16 at 17:54
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Set $u=t-2$ and $du=dt$

$$=\int\frac{du}{\sqrt u (u+2)^2}$$

Set $s=\sqrt u$ and $ds=\frac{du}{2\sqrt u}$

$$=2\int\frac{ds}{ (s^2+2)^2}$$

Set $s=\sqrt 2 \tan (p)$ so $p=\arctan (s/\sqrt 2)$

$$=\frac{1}{\sqrt 2}\int\cos^2(p)dp$$

Write as

$$=\frac{1}{\sqrt 2}\int \bigg(\frac 1 2\cos(2p)+\frac 1 2 \bigg)dp$$

$$=\frac{p}{2\sqrt 2}+\frac{\sin(2p)}{4\sqrt 2}+C$$

$$ \boxed{=\frac{2\sqrt{t-2} + \sqrt{2}t \arctan\left(\sqrt{\frac{t-2}{2}}\right)}{4t}+C}$$

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\begin{align} u & = \sqrt{t-2} \\[8pt] u^2 & = t-2 \\[8pt] u^2 + 2 & = t \\[8pt] 2u\,du & = dt \end{align} \begin{align} \int \frac{dt}{t^2\sqrt{t-2}} & = \int \frac{2u\,du}{(u^2+2)^2 u} = \int\frac{2\,du}{(u^2+2)^2} \end{align} Then let $u = \sqrt 2 \tan \theta$, $du= \sqrt 2 \sec^2\theta\,d\theta$, $u^2+2=2\sec^2\theta$. The integral becomes $$ \int \frac{2\sqrt 2 \sec^2\theta\,d\theta}{\sec^4\theta} = 2\sqrt2 \int\cos^2\theta\,d\theta, \quad\text{etc.} $$

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