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Antiderivative of Log

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tankia:
Wtf are you guys talking about *is intimidated* I don't know this stuff ><;

Freitag:

--- Quote from: "tankia" ---Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
--- End quote ---


Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.

Ie. dy/dx = ( 1/x ) --> y= log_e |x|

tankia:

--- Quote from: "Freitag" ---
--- Quote from: "tankia" ---Wtf are you guys talking about *is intimidated* I don't know this stuff ><;
--- End quote ---


Don't worry yourself too much.
Basically when you integrate a function with yields a log, (ie. dy/dx (1/x) --> y= log_e x), you put the brackets around the (in my example) x as modulus signs.

Ie. dy/dx = ( 1/x ) --> y= log_e |x|
--- End quote ---


Whyyy? Lol ....I'm so screwed for exams ><;

Odette:
Poor Tankia :( there there, i'm sure you'll be fine =]

Collin Li:

--- Quote from: "tankia" ---Whyyy? Lol ....I'm so screwed for exams ><;
--- End quote ---


dy/dx = 1/x, find y: for R\{0}

Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.

Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.

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