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







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  • $begingroup$
    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
    Jan 21 at 0:22










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    Jan 21 at 0:24










  • $begingroup$
    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
    Jan 21 at 0:24


















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
    Jan 21 at 0:22










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    Jan 21 at 0:24










  • $begingroup$
    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
    Jan 21 at 0:24
















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 Jan 21 at 0:25







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 Jan 21 at 0:19









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
    Jan 21 at 0:22










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    Jan 21 at 0:24










  • $begingroup$
    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
    Jan 21 at 0:24




















  • $begingroup$
    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
    Jan 21 at 0:22










  • $begingroup$
    you should include $dx$ and $du$
    $endgroup$
    – J. W. Tanner
    Jan 21 at 0:24










  • $begingroup$
    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
    Jan 21 at 0:24


















$begingroup$
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
Jan 21 at 0:22




$begingroup$
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
Jan 21 at 0:22












$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
Jan 21 at 0:24




$begingroup$
you should include $dx$ and $du$
$endgroup$
– J. W. Tanner
Jan 21 at 0:24












$begingroup$
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
Jan 21 at 0:24






$begingroup$
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
Jan 21 at 0:24












1 Answer
1






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oldest

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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
    Jan 21 at 0:28










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    Jan 21 at 0:31











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1 Answer
1






active

oldest

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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
    Jan 21 at 0:28










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    Jan 21 at 0:31
















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
    Jan 21 at 0:28










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    Jan 21 at 0:31














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 Jan 21 at 0:28

























answered Jan 21 at 0:27









Rhys HughesRhys Hughes

5,7841530




5,7841530












  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    Jan 21 at 0:28










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    Jan 21 at 0:31


















  • $begingroup$
    Thank you so much you are a life saver.
    $endgroup$
    – P.Lord
    Jan 21 at 0:28










  • $begingroup$
    You're most welcome. Thank you for the good question.
    $endgroup$
    – Rhys Hughes
    Jan 21 at 0:31
















$begingroup$
Thank you so much you are a life saver.
$endgroup$
– P.Lord
Jan 21 at 0:28




$begingroup$
Thank you so much you are a life saver.
$endgroup$
– P.Lord
Jan 21 at 0:28












$begingroup$
You're most welcome. Thank you for the good question.
$endgroup$
– Rhys Hughes
Jan 21 at 0:31




$begingroup$
You're most welcome. Thank you for the good question.
$endgroup$
– Rhys Hughes
Jan 21 at 0:31










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










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