Why is the integral of $1/x$ not $ln|ax|$?
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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
edited Nov 26 at 6:02
Robert Howard
1,9181822
asked Nov 26 at 3:36
Zach
12716
add a comment |
up vote
1
down vote
favorite
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
edited Nov 26 at 6:02
Robert Howard
1,9181822
asked Nov 26 at 3:36
Zach
12716
The a's cancel out. Try chain rule.
– Dude156
Nov 26 at 3:42
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited Nov 26 at 6:02
Robert Howard
1,9181822
asked Nov 26 at 3:36
Zach
12716
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
edited Nov 26 at 6:02
Robert Howard
1,9181822
asked Nov 26 at 3:36
Zach
12716
edited Nov 26 at 6:02
Robert Howard
1,9181822
asked Nov 26 at 3:36
Zach
12716
edited Nov 26 at 6:02
Robert Howard
1,9181822
edited Nov 26 at 6:02
Robert Howard
1,9181822
edited Nov 26 at 6:02
Robert Howard
1,9181822
1,9181822
asked Nov 26 at 3:36
Zach
12716
asked Nov 26 at 3:36
Zach
12716
asked Nov 26 at 3:36
Zach
12716
12716
The a's cancel out. Try chain rule.
– Dude156
Nov 26 at 3:42
add a comment |
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
add a comment |
1 Answer
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because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
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
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
2
down vote
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
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
add a comment |
up vote
2
down vote
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
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
add a comment |
up vote
2
down vote
up vote
2
down vote
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
answered Nov 26 at 3:41
Jimmy Sabater
1,827218
1,827218
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
add a comment |
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
add a comment |
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The a's cancel out. Try chain rule.
– Dude156
Nov 26 at 3:42