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






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.







$endgroup$












  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    1 hour 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
    1 hour 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






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.







$endgroup$












  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    1 hour 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
    1 hour 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






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.







$endgroup$




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




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 1 hour 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 1 hour 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
    1 hour ago










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










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












$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
1 hour ago




$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
1 hour 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
1 hour 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
1 hour 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
    1 hour ago










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










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










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    1 hour 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 1 hour ago

























answered 1 hour ago









Rhys HughesRhys Hughes

5,6461528




5,6461528












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










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


















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










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
















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




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












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




$begingroup$
You're most welcome. Thank you for the good question.
$endgroup$
– Rhys Hughes
1 hour 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

Усть-Каменогорск

Image
У этого термина существуют и другие значения, см. Усть-Каменогорск. Город Усть-Каменогорск каз. Өскемен Флаг Герб 49°57′ с. ш. 82°37′ в. д. H G Я O L Страна Казахстан   Казахстан Статус областной центр Область Восточно-Казахстанская Аким Жаксылык Омар [1] История и география Основан 1720 Город с 1868 Площадь 303,15 [2] км² Высота центра 283 ± 1 м Тип климата резко-континентальный Часовой пояс UTC+6 Население Население 329 090 [3]  человек ( 2018 ) Национальности русские 53,56 % казахи 42,99 % немцы 0,82 % украинцы 0,57 % татары 0,79 % белорусы 0,14 % азербайджанцы 0,23 % корейцы 0,20 % прочие 1,08 % -1 шт [4] Этнохороним усть-каменогорцы, усть-каменогорец Цифровые идентификаторы Телефонный код +7 (7232) Почтовый индекс F0***** [5] Автомобильный код 16 (ранее F) oskemen.kz Усть-Каменогорск Усть-Каменогорск ...
Read more

Халкинская богословская школа

Image
Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης Главное здание Халкинской школы Международное название Theological School of Halki Год основания 1844 Конфессия православие Церковь Константинопольская Ректор Елпидифор (Ламбриниадис) Расположение о. Хейбелиада, Турция Юридический адрес Ümit tepesi No 5, 34973 Heybeliada, Istanbul, Turkey Сайт www.greece.org  [[commons:Category:Halki seminary|Халкинская богословская школа тур. Rum Ortodoks Ruhban Okulu греч. Ιερά Θεολογική Σχολή της Χάλκης на Викискладе]] Ха́лкинская богосло́вская шко́ла (или Духовная семинария на Халки , тур. Rum Ortodoks Ruhban Okulu , греч. Ιερά Θεολογική Σχολή της Χάλκης ) — недействующее среднее специальное учебное заведение Константинопольской православной церкви, расположенное на территории Свято-Троицкого мужского монастыря на вершине холма Надежды (тур. Ümit Tepesi ) острова Хейбелиада, в Турции....
Read more

Высокополье (Харьковская область)

Image
У этого термина существуют и другие значения, см. Высокополье. Село Высокополье укр. Високопілля Герб 49°55′07″ с. ш. 35°25′13″ в. д. H G Я O L Страна Украина   Украина Область Харьковская Район Валковский Сельский совет Высокопольский История и география Основан 1670 Прежние названия Хмелевое Площадь 55,56 км² Часовой пояс UTC+2, летом UTC+3 Население Население 1957 человек ( 2001 ) Плотность 35,220 чел./км² Цифровые идентификаторы Телефонный код +380 5753 Почтовый индекс 63020 Автомобильный код AX, КХ / 21 КОАТУУ 6321281501 Высокополье Высокополье Высокополье  Аудио, фото и видео на Викискладе Высокопо́лье (укр. Високопілля ) — село, Высокопольский сельский совет, Валковский район, Харьковская область, Украина. Код КОАТУУ — 6321281501. Население по переписи 2001 г. составляет 1957 (866/1091 м/ж) человек. Является административным центр...
Read more