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    When using u substitution to evaluate an integral why is it possible to treat du/dx as a fraction and rearrange to replace dx?
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    you aren't treating it as a fraction any more than you do when you solve a differential equation, you are just changing the variable you are integrating with respect to. if you want a proper answer, you'll have to ask someone good i.e. nuodai
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    There are a few answers I can give to this question depending on the level you're working out. If you're doing A-level, the answer is "you just can". If you're doing anything higher, the answer really is that you can't except for notation's sake. So when we write u^2\, du = 2x\, dx we do so on the understanding that we're going to do something else with the equation, for example integrate both sides to give something like \displaystyle \int u^2\, du = \int 2x\, dx, or "divide through" by a differential to give something like u^2\dfrac{du}{dx} = 2x or u^2\dfrac{du}{dt} = 2x\dfrac{dx}{dt}. This is possible because rules like the chain rule make it possible.
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    (Original post by nuodai)
    There are a few answers I can give to this question depending on the level you're working out. If you're doing A-level, the answer is "you just can". If you're doing anything higher, the answer really is that you can't except for notation's sake. So when we write u^2\, du = 2x\, dx we do so on the understanding that we're going to do something else with the equation, for example integrate both sides to give something like \displaystyle \int u^2\, du = \int 2x\, dx, or "divide through" by a differential to give something like u^2\dfrac{du}{dx} = 2x or u^2\dfrac{du}{dt} = 2x\dfrac{dx}{dt}. This is possible because rules like the chain rule make it possible.
    thank you
 
 
 
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