Why is the integral of $1/x$ not $ln|ax|$?












1














Wouldn't any non-zero constant $a$ cause the derivative of $ln|ax|$ to simply be $1/x$? So shouldn't the integral of $1/x$ include the constant $a$? I am probably missing something basic here, thanks!



(EDIT: I just realized as I posted that with logarithm rules this $a$ value just goes to the $+C$ constant from integrating, so got!)










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  • The a's cancel out. Try chain rule.
    – Dude156
    Nov 26 at 3:42
















1














Wouldn't any non-zero constant $a$ cause the derivative of $ln|ax|$ to simply be $1/x$? So shouldn't the integral of $1/x$ include the constant $a$? I am probably missing something basic here, thanks!



(EDIT: I just realized as I posted that with logarithm rules this $a$ value just goes to the $+C$ constant from integrating, so got!)










share|cite|improve this question
























  • The a's cancel out. Try chain rule.
    – Dude156
    Nov 26 at 3:42














1












1








1







Wouldn't any non-zero constant $a$ cause the derivative of $ln|ax|$ to simply be $1/x$? So shouldn't the integral of $1/x$ include the constant $a$? I am probably missing something basic here, thanks!



(EDIT: I just realized as I posted that with logarithm rules this $a$ value just goes to the $+C$ constant from integrating, so got!)










share|cite|improve this question















Wouldn't any non-zero constant $a$ cause the derivative of $ln|ax|$ to simply be $1/x$? So shouldn't the integral of $1/x$ include the constant $a$? I am probably missing something basic here, thanks!



(EDIT: I just realized as I posted that with logarithm rules this $a$ value just goes to the $+C$ constant from integrating, so got!)







integration






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edited Nov 26 at 6:02









Robert Howard

1,9161822




1,9161822










asked Nov 26 at 3:36









Zach

12716




12716












  • The a's cancel out. Try chain rule.
    – Dude156
    Nov 26 at 3:42


















  • The a's cancel out. Try chain rule.
    – Dude156
    Nov 26 at 3:42
















The a's cancel out. Try chain rule.
– Dude156
Nov 26 at 3:42




The a's cancel out. Try chain rule.
– Dude156
Nov 26 at 3:42










1 Answer
1






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2














because of:



$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$






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  • 2




    Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
    – Zach
    Nov 26 at 3:51






  • 2




    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    – Jimmy Sabater
    Nov 26 at 4:13











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














because of:



$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$






share|cite|improve this answer

















  • 2




    Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
    – Zach
    Nov 26 at 3:51






  • 2




    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    – Jimmy Sabater
    Nov 26 at 4:13
















2














because of:



$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$






share|cite|improve this answer

















  • 2




    Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
    – Zach
    Nov 26 at 3:51






  • 2




    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    – Jimmy Sabater
    Nov 26 at 4:13














2












2








2






because of:



$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$






share|cite|improve this answer












because of:



$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 3:41









Jimmy Sabater

1,930219




1,930219








  • 2




    Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
    – Zach
    Nov 26 at 3:51






  • 2




    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    – Jimmy Sabater
    Nov 26 at 4:13














  • 2




    Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
    – Zach
    Nov 26 at 3:51






  • 2




    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    – Jimmy Sabater
    Nov 26 at 4:13








2




2




Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
– Zach
Nov 26 at 3:51




Yeah, I knew that. My question was why the anti-derivative didn't include an a, BECAUSE they cancel out. But I figured it it's part of the plus C.
– Zach
Nov 26 at 3:51




2




2




The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
– Jimmy Sabater
Nov 26 at 4:13




The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
– Jimmy Sabater
Nov 26 at 4:13


















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