How to calculate $int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx$ [on hold]
How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$
definite-integrals
edited Jan 6 at 20:52
amWhy
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asked Jan 6 at 17:25
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put on hold as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53
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How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$
definite-integrals
edited Jan 6 at 20:52
amWhy
192k28225439
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
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put on hold as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
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How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$
definite-integrals
edited Jan 6 at 20:52
amWhy
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asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
New contributor
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How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$
definite-integrals
definite-integrals
edited Jan 6 at 20:52
amWhy
192k28225439
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
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edited Jan 6 at 20:52
amWhy
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asked Jan 6 at 17:25
Adil AndersonAdil Anderson
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edited Jan 6 at 20:52
amWhy
192k28225439
edited Jan 6 at 20:52
amWhy
192k28225439
edited Jan 6 at 20:52
amWhy
192k28225439
192k28225439
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
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asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
62
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Adil Anderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Adil Anderson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by RRL, amWhy, A. Pongrácz, Gibbs, egreg Jan 6 at 22:53
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, amWhy, A. Pongrácz, Gibbs, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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Short answer: $I=1$.
Proof:
Note that
begin{align*}
I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
end{align*}
Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$
edited Jan 6 at 17:55
Bernard
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
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$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Short answer: $I=1$.
Proof:
Note that
begin{align*}
I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
end{align*}
Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$
edited Jan 6 at 17:55
Bernard
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
Short answer: $I=1$.
Proof:
Note that
begin{align*}
I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
end{align*}
Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$
edited Jan 6 at 17:55
Bernard
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
Short answer: $I=1$.
Proof:
Note that
begin{align*}
I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
end{align*}
Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$
edited Jan 6 at 17:55
Bernard
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Short answer: $I=1$.
Proof:
Note that
begin{align*}
I := int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx\
overset{text{substitute } v := 3-x}{=} - int_{-1}^1 -frac{sqrt{ln(6+v)}} {sqrt{ln(6+v)} + sqrt{ln(6-v)}} space dx\
overset{text{substitute } u := -v}{=} int_{-1}^1 -frac{sqrt{ln(6-u)}}{sqrt{ln(6+u)} + sqrt{ln(6-u)}} space dx
end{align*}
Thus, $2cdot I = displaystyleint_{-1}^1 frac{sqrt{ln(6-x)}+sqrt{ln(6+x)}}{sqrt{ln(6+x)} + sqrt{ln(6-x)}} space dx = int_{-1}^1 1 = 2$, i.e. $I = 1$
edited Jan 6 at 17:55
Bernard
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Jan 6 at 17:55
Bernard
118k639112
edited Jan 6 at 17:55
Bernard
118k639112
edited Jan 6 at 17:55
Bernard
118k639112
118k639112
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
46110
46110
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Maximilian Janisch is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
add a comment |
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
add a comment |
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
edited Jan 6 at 17:56
Bernard
118k639112
edited Jan 6 at 17:56
Bernard
118k639112
edited Jan 6 at 17:56
Bernard
118k639112
118k639112
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
2,039220
2,039220
add a comment |
add a comment |
