Stuck with Integration by Substitution












2












$begingroup$


I have a question where we need to find an integral using where "$u = 1+e^{x}$" for the equation "$int frac{e^{3x}}{1+e^{x}}dx$".



However when I substitute it I end up with "$int frac{(u-1)^{3}}{u}du$" instead of "$int frac{(u-1)^{2}}{u}du$" which is what I should be getting. Please help










share|cite|improve this question









New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
    $endgroup$
    – kccu
    2 hours ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    2 hours ago










  • $begingroup$
    Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
    $endgroup$
    – P.Lord
    2 hours ago


















2












$begingroup$


I have a question where we need to find an integral using where "$u = 1+e^{x}$" for the equation "$int frac{e^{3x}}{1+e^{x}}dx$".



However when I substitute it I end up with "$int frac{(u-1)^{3}}{u}du$" instead of "$int frac{(u-1)^{2}}{u}du$" which is what I should be getting. Please help










share|cite|improve this question









New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
    $endgroup$
    – kccu
    2 hours ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    2 hours ago










  • $begingroup$
    Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
    $endgroup$
    – P.Lord
    2 hours ago
















2












2








2





$begingroup$


I have a question where we need to find an integral using where "$u = 1+e^{x}$" for the equation "$int frac{e^{3x}}{1+e^{x}}dx$".



However when I substitute it I end up with "$int frac{(u-1)^{3}}{u}du$" instead of "$int frac{(u-1)^{2}}{u}du$" which is what I should be getting. Please help










share|cite|improve this question









New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have a question where we need to find an integral using where "$u = 1+e^{x}$" for the equation "$int frac{e^{3x}}{1+e^{x}}dx$".



However when I substitute it I end up with "$int frac{(u-1)^{3}}{u}du$" instead of "$int frac{(u-1)^{2}}{u}du$" which is what I should be getting. Please help







integration substitution






share|cite|improve this question









New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago







P.Lord













New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 3 hours ago









P.LordP.Lord

1113




1113




New contributor




P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






P.Lord is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
    $endgroup$
    – kccu
    2 hours ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    2 hours ago










  • $begingroup$
    Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
    $endgroup$
    – P.Lord
    2 hours ago




















  • $begingroup$
    You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
    $endgroup$
    – kccu
    2 hours ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    2 hours ago










  • $begingroup$
    Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
    $endgroup$
    – P.Lord
    2 hours ago


















$begingroup$
You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
$endgroup$
– kccu
2 hours ago




$begingroup$
You are correct, if $u=1+e^x$, then $e^x=u-1$ and $e^{3x}=(u-1)^3$. Is it possible whatever solutions you are looking at are incorrect?
$endgroup$
– kccu
2 hours ago












$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
2 hours ago




$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
2 hours ago












$begingroup$
Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
$endgroup$
– P.Lord
2 hours ago






$begingroup$
Have checked on two websites and both somehow have $e^{3x}=(u−1)^{2}$ and they get the same answer as the one on in the answers section.
$endgroup$
– P.Lord
2 hours ago












1 Answer
1






active

oldest

votes


















4












$begingroup$

$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
Also $e^{3x}=(u-1)^3$



Put it together and we have:



$$int frac{e^{3x}}{1+e^x} dx = int frac{(u-1)^3}{u}cdot frac{du}{u-1} =int frac{(u-1)^2}{u} du $$ as required



Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    2 hours ago










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    2 hours ago











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
});


}
});






P.Lord is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081312%2fstuck-with-integration-by-substitution%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









4












$begingroup$

$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
Also $e^{3x}=(u-1)^3$



Put it together and we have:



$$int frac{e^{3x}}{1+e^x} dx = int frac{(u-1)^3}{u}cdot frac{du}{u-1} =int frac{(u-1)^2}{u} du $$ as required



Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    2 hours ago










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    2 hours ago
















4












$begingroup$

$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
Also $e^{3x}=(u-1)^3$



Put it together and we have:



$$int frac{e^{3x}}{1+e^x} dx = int frac{(u-1)^3}{u}cdot frac{du}{u-1} =int frac{(u-1)^2}{u} du $$ as required



Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    2 hours ago










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    2 hours ago














4












4








4





$begingroup$

$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
Also $e^{3x}=(u-1)^3$



Put it together and we have:



$$int frac{e^{3x}}{1+e^x} dx = int frac{(u-1)^3}{u}cdot frac{du}{u-1} =int frac{(u-1)^2}{u} du $$ as required



Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.






share|cite|improve this answer











$endgroup$



$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
Also $e^{3x}=(u-1)^3$



Put it together and we have:



$$int frac{e^{3x}}{1+e^x} dx = int frac{(u-1)^3}{u}cdot frac{du}{u-1} =int frac{(u-1)^2}{u} du $$ as required



Your mistake was that you didnt substitute in $du$ in for $dx$ when you applied the u-sub.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 2 hours ago









Rhys HughesRhys Hughes

5,6461528




5,6461528












  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    2 hours ago










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    2 hours ago


















  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    2 hours ago










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    2 hours ago
















$begingroup$
Thank you so much you are a life saver.
$endgroup$
– P.Lord
2 hours ago




$begingroup$
Thank you so much you are a life saver.
$endgroup$
– P.Lord
2 hours ago












$begingroup$
You're most welcome. Thank you for the good question.
$endgroup$
– Rhys Hughes
2 hours ago




$begingroup$
You're most welcome. Thank you for the good question.
$endgroup$
– Rhys Hughes
2 hours ago










P.Lord is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















P.Lord is a new contributor. Be nice, and check out our Code of Conduct.













P.Lord is a new contributor. Be nice, and check out our Code of Conduct.












P.Lord is a new contributor. Be nice, and check out our Code of Conduct.
















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081312%2fstuck-with-integration-by-substitution%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

CARDNET

Boot-repair Failure: Unable to locate package grub-common:i386

濃尾地震