Evaluate $int (1+x)e^{2arctan x}dx$ [closed]
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Someone can help-me to calculate the integral
$int (1+x)e^{2arctan x}dx.$
I have try but without sucess.
calculus integration
edited Dec 3 '18 at 18:08
amWhy
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
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closed as off-topic by amWhy, RRL, GEdgar, Alexander Gruber♦ Dec 4 '18 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, GEdgar, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Someone can help-me to calculate the integral
$int (1+x)e^{2arctan x}dx.$
I have try but without sucess.
calculus integration
edited Dec 3 '18 at 18:08
amWhy
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
$endgroup$
closed as off-topic by amWhy, RRL, GEdgar, Alexander Gruber♦ Dec 4 '18 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, GEdgar, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
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– amWhy
Dec 3 '18 at 18:09
1
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
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– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
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– Doug M
Dec 3 '18 at 18:13
$begingroup$
Try the substitution $x=tan(u/2)$
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– clathratus
Dec 4 '18 at 4:29
add a comment |
$begingroup$
Someone can help-me to calculate the integral
$int (1+x)e^{2arctan x}dx.$
I have try but without sucess.
calculus integration
edited Dec 3 '18 at 18:08
amWhy
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
$endgroup$
Someone can help-me to calculate the integral
$int (1+x)e^{2arctan x}dx.$
I have try but without sucess.
calculus integration
calculus integration
edited Dec 3 '18 at 18:08
amWhy
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
edited Dec 3 '18 at 18:08
amWhy
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
edited Dec 3 '18 at 18:08
amWhy
1
edited Dec 3 '18 at 18:08
amWhy
1
edited Dec 3 '18 at 18:08
amWhy
1
1
asked Dec 3 '18 at 17:58
MotaKMotaK
73
asked Dec 3 '18 at 17:58
MotaKMotaK
73
asked Dec 3 '18 at 17:58
MotaKMotaK
73
73
closed as off-topic by amWhy, RRL, GEdgar, Alexander Gruber♦ Dec 4 '18 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, GEdgar, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, RRL, GEdgar, Alexander Gruber♦ Dec 4 '18 at 4:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, RRL, GEdgar, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
$endgroup$
– amWhy
Dec 3 '18 at 18:09
1
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
$endgroup$
– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
$endgroup$
– Doug M
Dec 3 '18 at 18:13
$begingroup$
Try the substitution $x=tan(u/2)$
$endgroup$
– clathratus
Dec 4 '18 at 4:29
add a comment |
1
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
$endgroup$
– amWhy
Dec 3 '18 at 18:09
1
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
$endgroup$
– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
$endgroup$
– Doug M
Dec 3 '18 at 18:13
$begingroup$
Try the substitution $x=tan(u/2)$
$endgroup$
– clathratus
Dec 4 '18 at 4:29
1
1
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
$endgroup$
– amWhy
Dec 3 '18 at 18:09
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
$endgroup$
– amWhy
Dec 3 '18 at 18:09
1
1
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
$endgroup$
– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
$endgroup$
– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
$endgroup$
– Doug M
Dec 3 '18 at 18:13
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
$endgroup$
– Doug M
Dec 3 '18 at 18:13
$begingroup$
Try the substitution $x=tan(u/2)$
$endgroup$
– clathratus
Dec 4 '18 at 4:29
$begingroup$
Try the substitution $x=tan(u/2)$
$endgroup$
– clathratus
Dec 4 '18 at 4:29
add a comment |
1 Answer
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begin{align}int xe^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int frac{x^2}{x^2+1}e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+int frac{e^{2arctan x}}{x^2+1}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+frac{1}{2}e^{2arctan x}end{align}
At this point, just ignore the intermediate steps and only consider the equation with the first integral and the last three terms we have found. Move the second term from the right side of the equation to the left side and you will find the original integral, as follows:
begin{align}&int (1+x)e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}+frac{1}{2}e^{2arctan x}+ C=frac{x^2+1}{2}e^{2arctan x}+Cend{align}
Needless to say, differentiate the result and you will find the integrand.
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In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
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– user621367
Dec 3 '18 at 23:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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begin{align}int xe^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int frac{x^2}{x^2+1}e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+int frac{e^{2arctan x}}{x^2+1}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+frac{1}{2}e^{2arctan x}end{align}
At this point, just ignore the intermediate steps and only consider the equation with the first integral and the last three terms we have found. Move the second term from the right side of the equation to the left side and you will find the original integral, as follows:
begin{align}&int (1+x)e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}+frac{1}{2}e^{2arctan x}+ C=frac{x^2+1}{2}e^{2arctan x}+Cend{align}
Needless to say, differentiate the result and you will find the integrand.
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$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
add a comment |
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begin{align}int xe^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int frac{x^2}{x^2+1}e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+int frac{e^{2arctan x}}{x^2+1}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+frac{1}{2}e^{2arctan x}end{align}
At this point, just ignore the intermediate steps and only consider the equation with the first integral and the last three terms we have found. Move the second term from the right side of the equation to the left side and you will find the original integral, as follows:
begin{align}&int (1+x)e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}+frac{1}{2}e^{2arctan x}+ C=frac{x^2+1}{2}e^{2arctan x}+Cend{align}
Needless to say, differentiate the result and you will find the integrand.
$endgroup$
$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
add a comment |
$begingroup$
begin{align}int xe^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int frac{x^2}{x^2+1}e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+int frac{e^{2arctan x}}{x^2+1}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+frac{1}{2}e^{2arctan x}end{align}
At this point, just ignore the intermediate steps and only consider the equation with the first integral and the last three terms we have found. Move the second term from the right side of the equation to the left side and you will find the original integral, as follows:
begin{align}&int (1+x)e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}+frac{1}{2}e^{2arctan x}+ C=frac{x^2+1}{2}e^{2arctan x}+Cend{align}
Needless to say, differentiate the result and you will find the integrand.
$endgroup$
begin{align}int xe^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int frac{x^2}{x^2+1}e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+int frac{e^{2arctan x}}{x^2+1}dx=frac{x^2}{2}e^{2arctan x}-int e^{2arctan x}dx+frac{1}{2}e^{2arctan x}end{align}
At this point, just ignore the intermediate steps and only consider the equation with the first integral and the last three terms we have found. Move the second term from the right side of the equation to the left side and you will find the original integral, as follows:
begin{align}&int (1+x)e^{2arctan x}dx=frac{x^2}{2}e^{2arctan x}+frac{1}{2}e^{2arctan x}+ C=frac{x^2+1}{2}e^{2arctan x}+Cend{align}
Needless to say, differentiate the result and you will find the integrand.
answered Dec 3 '18 at 22:51
user621367
$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
add a comment |
$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
$begingroup$
In the calculation I first used integration by parts and then the identity $frac{x^2}{x^2+1}=frac{x^2+1-1}{x^2+1}$.
$endgroup$
– user621367
Dec 3 '18 at 23:00
add a comment |
1
$begingroup$
What did you try? You said you have tried; what did you try? Can you add your work, please, even if you got stuck?
$endgroup$
– amWhy
Dec 3 '18 at 18:09
1
$begingroup$
$frac{x^2+1}{2};e^{2arctan(x)}$
$endgroup$
– GEdgar
Dec 3 '18 at 18:11
$begingroup$
$2arctan x = ln frac {x-i}{x+i}$
$endgroup$
– Doug M
Dec 3 '18 at 18:13
$begingroup$
Try the substitution $x=tan(u/2)$
$endgroup$
– clathratus
Dec 4 '18 at 4:29