anonymous
  • anonymous
\[\int\limits 1\div \sqrt{x}+x\]
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\int\limits 1\div \sqrt{x}+x
anonymous
  • anonymous
|dw:1333535677578:dw| should be an easy integral to do...
anonymous
  • anonymous
if that's what you mean...

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anonymous
  • anonymous
\[\Large \int\limits \frac{1}{\sqrt x}+x \;\;dx\] \[\Large \int\limits \left(x\right)^{-\frac12} +x\;\;\;dx\]
anonymous
  • anonymous
no its not that way
anonymous
  • anonymous
\[\Large \int\limits \left(x\right)^{-\frac12} \;dx +\int\limits x\;\;dx\] now use this formula... \[\LARGE \int\limits x^r \;dx \longrightarrow \int\limits \frac{x^{r+1}}{r+1} \;dx\]
anonymous
  • anonymous
then what way it is? ...
anonymous
  • anonymous
i gave up on latex... :)
anonymous
  • anonymous
its this way: \[\int\limits 1/ (\sqrt{x}+x)\]
anonymous
  • anonymous
...@dpaInc it's a bit hard ... but better to understand ;)
anonymous
  • anonymous
yep how do you know all the commands? experience?
anonymous
  • anonymous
\[\huge \int\limits \frac{1}{\sqrt {x+x}} \;\;dx\]
anonymous
  • anonymous
yes... I've been in some forums and I have experience as well (about 2 months LOL) ahaha....
anonymous
  • anonymous
\[\huge \int\limits \frac{1}{\sqrt x +x} \;\;dx\]
anonymous
  • anonymous
I made a mistake :P ...
Zarkon
  • Zarkon
\[\int\limits \frac{1}{\sqrt x +x} dx=\int\limits \frac{1}{\sqrt{x}(1+\sqrt x) }dx\]
anonymous
  • anonymous
@ Zarkon that how it looks
Zarkon
  • Zarkon
let \[u=1+\sqrt{x}\]
anonymous
  • anonymous
nice... u-substitution...
anonymous
  • anonymous
doesthe answer become 0.5 In(u) +c?
Zarkon
  • Zarkon
no
Zarkon
  • Zarkon
close
Zarkon
  • Zarkon
\[u=1+\sqrt{x}\] \[du=\frac{1}{2\sqrt{x}}dx\] \[2du=\frac{1}{\sqrt{x}}dx\]
anonymous
  • anonymous
oh ya i see the answer is 2 In (u) +c
anonymous
  • anonymous
thanks
Zarkon
  • Zarkon
don't forget to replace \(u\) by \(1+\sqrt{x}\)
anonymous
  • anonymous
yes thanks

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