September 17th, 2016, 03:34 AM  #11 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
it' of de form $\int u^\alpha.u' f 
September 17th, 2016, 03:36 AM  #12 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
with x=1/t it becomes of the form $\int u^a u' f$ 
September 17th, 2016, 08:12 AM  #13 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
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. 
September 17th, 2016, 09:30 PM  #14 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,981 Thanks: 1166 Math Focus: Elementary mathematics and beyond 
$\displaystyle \int(x^2+1)^{3/2}\,dx=\int\frac{1}{(x^2+1)\sqrt{x^2+1}}\,dx= \int\frac{x} {x(x^2+1)\sqrt{x^2+1}}\,dx$ $$u=\frac{x}{\sqrt{x^2+1}},\quad du=(x^2+1)^{3/2}\,dx$$ ... etc ... 
September 17th, 2016, 11:01 PM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
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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