How to calculate $int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}} space dx$ [closed]
$begingroup$
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
Namaste
1
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
$endgroup$
closed as off-topic by RRL, Namaste, 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, Namaste, A. Pongrácz, Gibbs, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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
Namaste
1
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
$endgroup$
closed as off-topic by RRL, Namaste, 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, Namaste, A. Pongrácz, Gibbs, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
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
Namaste
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
edited Jan 6 at 20:52
Namaste
1
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
edited Jan 6 at 20:52
Namaste
1
asked Jan 6 at 17:25
Adil AndersonAdil Anderson
62
edited Jan 6 at 20:52
Namaste
1
edited Jan 6 at 20:52
Namaste
1
edited Jan 6 at 20:52
Namaste
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, Namaste, 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, Namaste, 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, Namaste, 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, Namaste, A. Pongrácz, Gibbs, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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. $I = 1$
edited Jan 6 at 17:55
Bernard
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
$endgroup$
add a comment |
$begingroup$
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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. $I = 1$
edited Jan 6 at 17:55
Bernard
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
$endgroup$
add a comment |
$begingroup$
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. $I = 1$
edited Jan 6 at 17:55
Bernard
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
$endgroup$
add a comment |
$begingroup$
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. $I = 1$
edited Jan 6 at 17:55
Bernard
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
$endgroup$
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. $I = 1$
edited Jan 6 at 17:55
Bernard
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
edited Jan 6 at 17:55
Bernard
124k741117
edited Jan 6 at 17:55
Bernard
124k741117
edited Jan 6 at 17:55
Bernard
124k741117
124k741117
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
answered Jan 6 at 17:28
Maximilian JanischMaximilian Janisch
44612
44612
add a comment |
add a comment |
$begingroup$
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
add a comment |
$begingroup$
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
add a comment |
$begingroup$
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
$$mathbf I =int_2^4 frac{sqrt{ln(9-x)}}{sqrt{ln(9-x)} + sqrt{ln(3+x)}}mathrm{dx}=int_2^4 frac{sqrt{ln(9-(4+2-x))}}{sqrt{ln(9-(4+2-x))} + sqrt{ln(3+(4+2-x))}}mathrm{dx} qquadtext{(Why?)}$$
So
$mathbf I =displaystyleint_2^4 frac{sqrt{ln(3+x)}}{sqrt{ln(3+x)} + sqrt{ln(9-x)}}mathrm{dx}$.
Then $2mathbf I =int_2^4 mathrm{dx}=2$.
edited Jan 6 at 17:56
Bernard
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
edited Jan 6 at 17:56
Bernard
124k741117
edited Jan 6 at 17:56
Bernard
124k741117
edited Jan 6 at 17:56
Bernard
124k741117
124k741117
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
answered Jan 6 at 17:33
Thomas ShelbyThomas Shelby
4,7362727
4,7362727
add a comment |
add a comment |
