Evaluate $int frac e^x sqrt e^x + e^2x text dx.$ The Next CEO of Stack OverflowIntegration of $ int frac1x^2sqrtx^2 + 9dx $ using trigonometric substitutionTrigonometric Substitution Evaluate ∫(x^3)/(sqrt(9-4x^2))dxEvaluate $int(xsqrt1-x^4)dx$$intfracsqrtln xx dx$Integrate $int left( frac1-x1+x right)^frac32dx$Evaluate $intdfracdaasqrta+1$Solving $intfracsqrt4+xx, text dx$Integration help $frac1x^2 - x + 1$?Find $int fracarcsin sqrtx(1+x)sqrtxrm dx$.Does my book lie about $int fracarccos(fracx2)sqrt4-x^2dx$?
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Evaluate $int frac e^x sqrt e^x + e^2x text dx.$
The Next CEO of Stack OverflowIntegration of $ int frac1x^2sqrtx^2 + 9dx $ using trigonometric substitutionTrigonometric Substitution Evaluate ∫(x^3)/(sqrt(9-4x^2))dxEvaluate $int(xsqrt1-x^4)dx$$intfracsqrtln xx dx$Integrate $int left( frac1-x1+x right)^frac32dx$Evaluate $intdfracdaasqrta+1$Solving $intfracsqrt4+xx, text dx$Integration help $frac1x^2 - x + 1$?Find $int fracarcsin sqrtx(1+x)sqrtxrm dx$.Does my book lie about $int fracarccos(fracx2)sqrt4-x^2dx$?
$begingroup$
I started out with $U$-sub, making $e^x$ as $U.$ Replaced everything with $U$ and now I am stuck!
calculus integration indefinite-integrals
edited Mar 28 at 4:12
Dbchatto67
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I started out with $U$-sub, making $e^x$ as $U.$ Replaced everything with $U$ and now I am stuck!
calculus integration indefinite-integrals
edited Mar 28 at 4:12
Dbchatto67
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
1
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01
add a comment |
$begingroup$
I started out with $U$-sub, making $e^x$ as $U.$ Replaced everything with $U$ and now I am stuck!
calculus integration indefinite-integrals
edited Mar 28 at 4:12
Dbchatto67
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I started out with $U$-sub, making $e^x$ as $U.$ Replaced everything with $U$ and now I am stuck!
calculus integration indefinite-integrals
calculus integration indefinite-integrals
edited Mar 28 at 4:12
Dbchatto67
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 4:12
Dbchatto67
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 4:12
Dbchatto67
2,452422
edited Mar 28 at 4:12
Dbchatto67
2,452422
edited Mar 28 at 4:12
Dbchatto67
2,452422
2,452422
asked Mar 28 at 3:30
GeorgeGeorge
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 3:30
GeorgeGeorge
9
asked Mar 28 at 3:30
GeorgeGeorge
9
9
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
George is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
1
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01
add a comment |
5
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
1
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01
5
5
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
1
1
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint $colon$ Observe that $$frac e^x sqrt e^x + e^2x = frac e^frac x 2 sqrt 1 + e^x.$$ Take the substitution $frac x 2 = z.$ Then $ dx =2 dz.$ So the integral becomes $$2 int frac e^z sqrt 1 + e^2z dz.$$ Now take the substitution $e^z = u.$ So $e^zdz = du.$ Then the integral reduces to $$2 int frac du sqrt 1+u^2.$$ At last take the substitution $u = tan theta.$ Then $ du = sec^2 theta d theta.$ So the integral further reduces to $$2 int sec theta d theta.$$ Can you proceed now?
edited Mar 28 at 5:09
mouthetics
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
$endgroup$
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
add a comment |
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$begingroup$
Hint $colon$ Observe that $$frac e^x sqrt e^x + e^2x = frac e^frac x 2 sqrt 1 + e^x.$$ Take the substitution $frac x 2 = z.$ Then $ dx =2 dz.$ So the integral becomes $$2 int frac e^z sqrt 1 + e^2z dz.$$ Now take the substitution $e^z = u.$ So $e^zdz = du.$ Then the integral reduces to $$2 int frac du sqrt 1+u^2.$$ At last take the substitution $u = tan theta.$ Then $ du = sec^2 theta d theta.$ So the integral further reduces to $$2 int sec theta d theta.$$ Can you proceed now?
edited Mar 28 at 5:09
mouthetics
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
$endgroup$
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
add a comment |
$begingroup$
Hint $colon$ Observe that $$frac e^x sqrt e^x + e^2x = frac e^frac x 2 sqrt 1 + e^x.$$ Take the substitution $frac x 2 = z.$ Then $ dx =2 dz.$ So the integral becomes $$2 int frac e^z sqrt 1 + e^2z dz.$$ Now take the substitution $e^z = u.$ So $e^zdz = du.$ Then the integral reduces to $$2 int frac du sqrt 1+u^2.$$ At last take the substitution $u = tan theta.$ Then $ du = sec^2 theta d theta.$ So the integral further reduces to $$2 int sec theta d theta.$$ Can you proceed now?
edited Mar 28 at 5:09
mouthetics
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
$endgroup$
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
add a comment |
$begingroup$
Hint $colon$ Observe that $$frac e^x sqrt e^x + e^2x = frac e^frac x 2 sqrt 1 + e^x.$$ Take the substitution $frac x 2 = z.$ Then $ dx =2 dz.$ So the integral becomes $$2 int frac e^z sqrt 1 + e^2z dz.$$ Now take the substitution $e^z = u.$ So $e^zdz = du.$ Then the integral reduces to $$2 int frac du sqrt 1+u^2.$$ At last take the substitution $u = tan theta.$ Then $ du = sec^2 theta d theta.$ So the integral further reduces to $$2 int sec theta d theta.$$ Can you proceed now?
edited Mar 28 at 5:09
mouthetics
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
$endgroup$
Hint $colon$ Observe that $$frac e^x sqrt e^x + e^2x = frac e^frac x 2 sqrt 1 + e^x.$$ Take the substitution $frac x 2 = z.$ Then $ dx =2 dz.$ So the integral becomes $$2 int frac e^z sqrt 1 + e^2z dz.$$ Now take the substitution $e^z = u.$ So $e^zdz = du.$ Then the integral reduces to $$2 int frac du sqrt 1+u^2.$$ At last take the substitution $u = tan theta.$ Then $ du = sec^2 theta d theta.$ So the integral further reduces to $$2 int sec theta d theta.$$ Can you proceed now?
edited Mar 28 at 5:09
mouthetics
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
edited Mar 28 at 5:09
mouthetics
52137
edited Mar 28 at 5:09
mouthetics
52137
edited Mar 28 at 5:09
mouthetics
52137
52137
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
answered Mar 28 at 4:25
Dbchatto67Dbchatto67
2,452422
2,452422
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
add a comment |
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
How did you get the first statement? I am not quite seeing the steps you took> Thanks!
$endgroup$
– George
Mar 28 at 4:31
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
$begingroup$
Divide both numerator and denominator by $e^frac x 2.$
$endgroup$
– Dbchatto67
Mar 28 at 4:35
add a comment |
George is a new contributor. Be nice, and check out our Code of Conduct.
George is a new contributor. Be nice, and check out our Code of Conduct.
George is a new contributor. Be nice, and check out our Code of Conduct.
George is a new contributor. Be nice, and check out our Code of Conduct.
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5
$begingroup$
This website is not for us to solve homework for you. What have you tried so far?
$endgroup$
– Victoria M
Mar 28 at 3:33
1
$begingroup$
Hint: $e^2x=(e^x)^2.$ Now complete the square.
$endgroup$
– Sean Roberson
Mar 28 at 3:34
$begingroup$
@ Sean Roberson, I have just done that getting radical (u+ 1/2)^2 -1/4, but what would I need to do next? Would trig substitution be the right choice?
$endgroup$
– George
Mar 28 at 4:01