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Mathematics 8 Online
OpenStudy (anonymous):

integral (3x^2 / (squareroot of x^3 +1) )dx =

OpenStudy (misty1212):

HI!

OpenStudy (misty1212):

this is a u sub for sure

OpenStudy (misty1212):

put \(u=x^3+1, du =3x^2dx\) and you get \[\int \frac{du}{\sqrt{u}}\] right away

OpenStudy (misty1212):

you okay after that? write is as \[\int u^{-\frac{1}{2}}du\] and use the power rule, only the other way \[\int u^n=\frac{u^{n+1}}{n+1}\]

OpenStudy (anonymous):

okay keep going

OpenStudy (misty1212):

lol no you keep going let me know what you get

OpenStudy (anonymous):

1/2 ( x^3 +1) ^ -1/2 (3x^2) dx

OpenStudy (anonymous):

then 2du = all that minus the 1/2

myininaya (myininaya):

where does that 1/2 in front come from ?

OpenStudy (misty1212):

i guess i did not help much

OpenStudy (misty1212):

start here \[\int u^{-\frac{1}{2}}du\]

OpenStudy (anonymous):

okay yes i go that far

myininaya (myininaya):

|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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