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September 17th, 2016, 03:34 AM   #11
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it' of de form

$\int u^\alpha.u' f
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September 17th, 2016, 03:36 AM   #12
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with x=1/t it becomes of the form

$\int u^a u' f$
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September 17th, 2016, 08:12 AM   #13
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don't take into account posts 11 and 12.

with the substitution $x=\frac{1}{t}$, your integral becomes of the form

$\int u^a u'$

Last edited by abdallahhammam; September 17th, 2016 at 08:15 AM.
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September 17th, 2016, 09:30 PM   #14
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$\displaystyle \int(x^2+1)^{-3/2}\,dx=\int\frac{1}{(x^2+1)\sqrt{x^2+1}}\,dx= \int\frac{x} {x(x^2+1)\sqrt{x^2+1}}\,dx$

$$u=\frac{x}{\sqrt{x^2+1}},\quad du=(x^2+1)^{-3/2}\,dx$$

... etc ...

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September 17th, 2016, 11:01 PM   #15
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I like the $\frac1t$ idea although it essentially becomes the single substitution that I used.

The one I came up with was
$$\begin{aligned}
\int \left(1+x^2\right)^{-\frac32} \,\mathrm dx &= \int {1 \over \left(\sqrt{1+x^2}\right)^3} \,\mathrm dx \\
&= \int {1 \over \left(\sqrt{x^2\left(\frac1{x^2}+1\right)}\right)^ 3} \,\mathrm dx \\
&= \int {1 \over \left(x\sqrt{\frac1{x^2}+1}\right)^3} \,\mathrm dx \\
&= \int {1 \over x^3\left(\sqrt{\frac1{x^2}+1}\right)^3} \,\mathrm dx \\
&= \int x^{-3}\left(\tfrac1{x^2}+1\right)^{-\frac32} \,\mathrm dx \\
\end{aligned}$$
And then $u = 1 + x^{-2}$.
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