How to calculate $int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx$ [closed]

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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

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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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closed 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

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edited Jan 6 at 20:52

amWhy

1

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

$endgroup$

closed 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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How to calculate $$int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} spacemathrm{d}x$$

definite-integrals

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edited Jan 6 at 20:52

amWhy

1

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

$endgroup$

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

share|cite|improve this question

edited Jan 6 at 20:52

amWhy

1

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

share|cite|improve this question

edited Jan 6 at 20:52

amWhy

1

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

share|cite|improve this question

share|cite|improve this question

edited Jan 6 at 20:52

amWhy

1

edited Jan 6 at 20:52

amWhy

1

edited Jan 6 at 20:52

amWhy

1

1

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

asked Jan 6 at 17:25

Adil AndersonAdil Anderson

62

62

closed 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.

closed 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
2

active

oldest

votes

2

$begingroup$

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. $I = 1$

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edited Jan 6 at 17:55

Bernard

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

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1

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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 $2mathbf I =int_2^4 mathrm{dx}=2$.

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edited Jan 6 at 17:56

Bernard

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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

2

$begingroup$

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. $I = 1$

share|cite|improve this answer

edited Jan 6 at 17:55

Bernard

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

$endgroup$

add a comment | 

2

$begingroup$

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. $I = 1$

share|cite|improve this answer

edited Jan 6 at 17:55

Bernard

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

$endgroup$

add a comment | 

2

2

2

$begingroup$

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. $I = 1$

share|cite|improve this answer

edited Jan 6 at 17:55

Bernard

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

$endgroup$

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. $I = 1$

share|cite|improve this answer

edited Jan 6 at 17:55

Bernard

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

share|cite|improve this answer

share|cite|improve this answer

edited Jan 6 at 17:55

Bernard

119k740113

edited Jan 6 at 17:55

Bernard

119k740113

edited Jan 6 at 17:55

Bernard

119k740113

119k740113

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

answered Jan 6 at 17:28

Maximilian JanischMaximilian Janisch

44110

44110

add a comment | 

add a comment | 

1

$begingroup$

$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.

Then $2mathbf I =int_2^4 mathrm{dx}=2$.

share|cite|improve this answer

edited Jan 6 at 17:56

Bernard

119k740113

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

$endgroup$

add a comment | 

1

$begingroup$

$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.

Then $2mathbf I =int_2^4 mathrm{dx}=2$.

share|cite|improve this answer

edited Jan 6 at 17:56

Bernard

119k740113

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

$endgroup$

add a comment | 

1

1

1

$begingroup$

$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.

Then $2mathbf I =int_2^4 mathrm{dx}=2$.

share|cite|improve this answer

edited Jan 6 at 17:56

Bernard

119k740113

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

$endgroup$

$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.

Then $2mathbf I =int_2^4 mathrm{dx}=2$.

share|cite|improve this answer

edited Jan 6 at 17:56

Bernard

119k740113

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

share|cite|improve this answer

share|cite|improve this answer

edited Jan 6 at 17:56

Bernard

119k740113

edited Jan 6 at 17:56

Bernard

119k740113

edited Jan 6 at 17:56

Bernard

119k740113

119k740113

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

answered Jan 6 at 17:33

Thomas ShelbyThomas Shelby

2,550421

2,550421

add a comment | 

add a comment | 

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