So, yes, this matches the form of ∫ (1/u) du = ln |u| + C Now let us see if we can put this in the form of 1/u du (derivative of that function) then the integral is So, the key to solving these problems is to pick something to call u so that you have a function of u that you know how to integrate multiplied by the derivative of whatever you called u. If you don't have that derivative, then you have to solve the integral some other way. (some function of u)(derivative of u - note, of u, not the whole function) You must have the derivative of whatever you are calling u in order to use the form. Let me try to explain the same thing in a way that might be more clear:Īll of the standard forms for integrals involve the following: I think it is a matter of f(x) in the example being used in more than one way.
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