Why is the integral of $1/x$ not $ln|ax|$?
$begingroup$
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
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
$endgroup$
add a comment |
$begingroup$
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
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
$endgroup$
$begingroup$
The a's cancel out. Try chain rule.
$endgroup$
– Dude156
Nov 26 '18 at 3:42
add a comment |
$begingroup$
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
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
$endgroup$
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
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
edited Nov 26 '18 at 6:02
Robert Howard
1,9161822
1,9161822
asked Nov 26 '18 at 3:36
ZachZach
12716
asked Nov 26 '18 at 3:36
ZachZach
12716
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
$endgroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013790%2fwhy-is-the-integral-of-1-x-not-lnax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
$endgroup$
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
add a comment |
$begingroup$
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
$endgroup$
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
add a comment |
$begingroup$
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
$endgroup$
because of:
$$ dfrac{ d log ax }{dx} = frac{ (ax)' }{ax} = frac{a}{ax} = frac{1}{x } $$
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
answered Nov 26 '18 at 3:41
Jimmy SabaterJimmy Sabater
2,225319
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
add a comment |
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013790%2fwhy-is-the-integral-of-1-x-not-lnax%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The a's cancel out. Try chain rule.
$endgroup$
– Dude156
Nov 26 '18 at 3:42