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












1












$begingroup$


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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$endgroup$












  • $begingroup$
    The a's cancel out. Try chain rule.
    $endgroup$
    – Dude156
    Nov 26 '18 at 3:42
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    The a's cancel out. Try chain rule.
    $endgroup$
    – Dude156
    Nov 26 '18 at 3:42














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 6:02









Robert Howard

1,9161822




1,9161822










asked Nov 26 '18 at 3:36









ZachZach

12716




12716












  • $begingroup$
    The a's cancel out. Try chain rule.
    $endgroup$
    – Dude156
    Nov 26 '18 at 3:42


















  • $begingroup$
    The a's cancel out. Try chain rule.
    $endgroup$
    – Dude156
    Nov 26 '18 at 3:42
















$begingroup$
The a's cancel out. Try chain rule.
$endgroup$
– Dude156
Nov 26 '18 at 3:42




$begingroup$
The a's cancel out. Try chain rule.
$endgroup$
– Dude156
Nov 26 '18 at 3:42










1 Answer
1






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oldest

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2












$begingroup$

because of:



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






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    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.
    $endgroup$
    – Zach
    Nov 26 '18 at 3:51






  • 2




    $begingroup$
    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    $endgroup$
    – Jimmy Sabater
    Nov 26 '18 at 4:13











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

because of:



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






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    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.
    $endgroup$
    – Zach
    Nov 26 '18 at 3:51






  • 2




    $begingroup$
    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    $endgroup$
    – Jimmy Sabater
    Nov 26 '18 at 4:13
















2












$begingroup$

because of:



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






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    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.
    $endgroup$
    – Zach
    Nov 26 '18 at 3:51






  • 2




    $begingroup$
    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    $endgroup$
    – Jimmy Sabater
    Nov 26 '18 at 4:13














2












2








2





$begingroup$

because of:



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






share|cite|improve this answer









$endgroup$



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 '18 at 3:41









Jimmy SabaterJimmy Sabater

2,225319




2,225319








  • 2




    $begingroup$
    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.
    $endgroup$
    – Zach
    Nov 26 '18 at 3:51






  • 2




    $begingroup$
    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    $endgroup$
    – Jimmy Sabater
    Nov 26 '18 at 4:13














  • 2




    $begingroup$
    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.
    $endgroup$
    – Zach
    Nov 26 '18 at 3:51






  • 2




    $begingroup$
    The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
    $endgroup$
    – Jimmy Sabater
    Nov 26 '18 at 4:13








2




2




$begingroup$
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.
$endgroup$
– Zach
Nov 26 '18 at 3:51




$begingroup$
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.
$endgroup$
– Zach
Nov 26 '18 at 3:51




2




2




$begingroup$
The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
$endgroup$
– Jimmy Sabater
Nov 26 '18 at 4:13




$begingroup$
The constant of integration $C$ eats it out because remember $log(ax) = log a + log x $ and $log a $ is a constant
$endgroup$
– Jimmy Sabater
Nov 26 '18 at 4:13


















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