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integral (3x^2 / (squareroot of x^3 +1) )dx =
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HI!
this is a u sub for sure
put \(u=x^3+1, du =3x^2dx\) and you get \[\int \frac{du}{\sqrt{u}}\] right away
you okay after that? write is as \[\int u^{-\frac{1}{2}}du\] and use the power rule, only the other way \[\int u^n=\frac{u^{n+1}}{n+1}\]
okay keep going
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lol no you keep going let me know what you get
1/2 ( x^3 +1) ^ -1/2 (3x^2) dx
then 2du = all that minus the 1/2
where does that 1/2 in front come from ?
i guess i did not help much
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start here \[\int u^{-\frac{1}{2}}du\]
okay yes i go that far
|dw:1421034275187:dw| the derivative of the inside of that bottom square root function is on top if u=x^3+1 then du/dx=3x^2 so du=3x^2 dx so ...
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