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Hints to find a certain integral
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How can I find $intsqrtx/(x^2+1),dx$?
I tried by substitution and by integration by parts but it doesn't seem to work. I only need hints, not answers.
calculus integration
edited Jul 22 at 20:53
Michael Hardy
204k23186462
asked Jul 22 at 20:15
user573497
2009
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up vote
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How can I find $intsqrtx/(x^2+1),dx$?
I tried by substitution and by integration by parts but it doesn't seem to work. I only need hints, not answers.
calculus integration
edited Jul 22 at 20:53
Michael Hardy
204k23186462
asked Jul 22 at 20:15
user573497
2009
2
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
1
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56
add a comment |Â
up vote
2
down vote
favorite
2
up vote
2
down vote
favorite
2
2
How can I find $intsqrtx/(x^2+1),dx$?
I tried by substitution and by integration by parts but it doesn't seem to work. I only need hints, not answers.
calculus integration
edited Jul 22 at 20:53
Michael Hardy
204k23186462
asked Jul 22 at 20:15
user573497
2009
How can I find $intsqrtx/(x^2+1),dx$?
I tried by substitution and by integration by parts but it doesn't seem to work. I only need hints, not answers.
calculus integration
edited Jul 22 at 20:53
Michael Hardy
204k23186462
asked Jul 22 at 20:15
user573497
2009
edited Jul 22 at 20:53
Michael Hardy
204k23186462
edited Jul 22 at 20:53
Michael Hardy
204k23186462
edited Jul 22 at 20:53
Michael Hardy
204k23186462
204k23186462
asked Jul 22 at 20:15
user573497
2009
asked Jul 22 at 20:15
user573497
2009
asked Jul 22 at 20:15
user573497
2009
2009
2
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
1
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56
add a comment |Â
2
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
1
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56
2
2
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
1
1
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56
add a comment |Â
1 Answer
1
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votes
up vote
2
down vote
accepted
Try taking
$$u=sqrtxtext and du=dfrac12sqrtx,dx$$
Then you get
$$2intfracu^2u^4+1 , du$$
Can you take it from here?
edited Jul 22 at 20:53
Michael Hardy
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Try taking
$$u=sqrtxtext and du=dfrac12sqrtx,dx$$
Then you get
$$2intfracu^2u^4+1 , du$$
Can you take it from here?
edited Jul 22 at 20:53
Michael Hardy
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
add a comment |Â
up vote
2
down vote
accepted
Try taking
$$u=sqrtxtext and du=dfrac12sqrtx,dx$$
Then you get
$$2intfracu^2u^4+1 , du$$
Can you take it from here?
edited Jul 22 at 20:53
Michael Hardy
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Try taking
$$u=sqrtxtext and du=dfrac12sqrtx,dx$$
Then you get
$$2intfracu^2u^4+1 , du$$
Can you take it from here?
edited Jul 22 at 20:53
Michael Hardy
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
Try taking
$$u=sqrtxtext and du=dfrac12sqrtx,dx$$
Then you get
$$2intfracu^2u^4+1 , du$$
Can you take it from here?
edited Jul 22 at 20:53
Michael Hardy
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
edited Jul 22 at 20:53
Michael Hardy
204k23186462
edited Jul 22 at 20:53
Michael Hardy
204k23186462
edited Jul 22 at 20:53
Michael Hardy
204k23186462
204k23186462
answered Jul 22 at 20:30
Key Flex
4,278423
answered Jul 22 at 20:30
Key Flex
4,278423
answered Jul 22 at 20:30
Key Flex
4,278423
4,278423
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
add a comment |Â
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
That seems to be alright. Thank you for the help! :)
– user573497
Jul 26 at 4:36
add a comment |Â
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2
Substitute $x=t^2$ begineqnarray* 2 int fract^2(t^2+sqrt2t+1)(t^2-sqrt2t+1) dt. endeqnarray* Now do partial fractions ... good luck $ddot smile$
– Donald Splutterwit
Jul 22 at 20:31
1
This comment may or may not be a better answer than the one answer that's been posted so far, depending on the needs of those asking the question, but it seems to me that factoring $t^4+1$ in that way is something that students learning this sort of thing for the first time often don't know how to do.
– Michael Hardy
Jul 22 at 20:56