Mixing
Stuck with Integration by Substitution
$begingroup$
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
asked Jan 21 at 0:19
P.LordP.Lord
1113
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add a comment |
$begingroup$
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
asked Jan 21 at 0:19
P.LordP.Lord
1113
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 21 at 0:19
P.LordP.Lord
1113
$endgroup$
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
asked Jan 21 at 0:19
P.LordP.Lord
1113
asked Jan 21 at 0:19
P.LordP.Lord
1113
edited Jan 21 at 0:25
P.Lord
asked Jan 21 at 0:19
P.LordP.Lord
1113
asked Jan 21 at 0:19
P.LordP.Lord
1113
asked Jan 21 at 0:19
P.LordP.Lord
1113
1113
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
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.
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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votes
$begingroup$
$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
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.
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
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.
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
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.
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$$u=1+e^x to du= e^x dx to dx =frac{du}{u-1}$$
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.
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
edited Jan 21 at 0:28
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
answered Jan 21 at 0:27
Rhys HughesRhys Hughes
6,9571530
6,9571530
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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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