Finding an integral using a table?Integral of $cosleft(frac1xright), dx$How to solve this indefinite integral...
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Finding an integral using a table?
Integral of $cosleft(frac1xright), dx$How to solve this indefinite integral using integral substitution?Two solutions for the same integral question, which approach is correct/better to solve?Solving integral without simplifying equationFinding the integral of $x^2sqrt[3]{1-x}$How to find the value of this indefinite integral?Finding double integral of this region using polar coordinates?Solving Integral with Symbolic Computation (Sympy), Division and Tricky LimitsProving Table of Integral Integral (Trigonometric Substitution)Using a table of integrals for solving these integrals
$begingroup$
Am I correct for pattern matching this integral?
I have
$$int frac{sqrt{9x^2+4}}{x^2}dx$$
Does this pattern match with:
$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$
If I factor out the 9, I get
$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$
I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited 1 hour ago
clathratus
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
$endgroup$
add a comment |
$begingroup$
Am I correct for pattern matching this integral?
I have
$$int frac{sqrt{9x^2+4}}{x^2}dx$$
Does this pattern match with:
$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$
If I factor out the 9, I get
$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$
I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited 1 hour ago
clathratus
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
$endgroup$
1
$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
1
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago
add a comment |
$begingroup$
Am I correct for pattern matching this integral?
I have
$$int frac{sqrt{9x^2+4}}{x^2}dx$$
Does this pattern match with:
$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$
If I factor out the 9, I get
$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$
I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited 1 hour ago
clathratus
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
$endgroup$
Am I correct for pattern matching this integral?
I have
$$int frac{sqrt{9x^2+4}}{x^2}dx$$
Does this pattern match with:
$$int frac{sqrt{a^2 + x^2}}{x^2}dx = -frac{a^2 + x^2}{x} + ln(x + sqrt{a^2 + x^2}) + c$$
If I factor out the 9, I get
$$= 3 int frac{sqrt{x^2 + frac{4}{9}}}{x^2}$$
with $a = frac{2}{3}$
I get:
$$3 left( - frac{sqrt{frac{4}{9}+x^2}}{x} + lnleft(x+sqrt{frac{4}{9}+x^2}right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
integration
edited 1 hour ago
clathratus
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
edited 1 hour ago
clathratus
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
edited 1 hour ago
clathratus
4,745337
edited 1 hour ago
clathratus
4,745337
edited 1 hour ago
clathratus
4,745337
4,745337
asked 3 hours ago
Jwan622Jwan622
2,20611632
asked 3 hours ago
Jwan622Jwan622
2,20611632
asked 3 hours ago
Jwan622Jwan622
2,20611632
2,20611632
1
$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
1
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago
add a comment |
1
$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
1
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago
1
1
$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
1
1
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
answered 3 hours ago
DeepakDeepak
17.3k11537
$endgroup$
add a comment |
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$begingroup$
You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
answered 3 hours ago
DeepakDeepak
17.3k11537
$endgroup$
add a comment |
$begingroup$
You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
answered 3 hours ago
DeepakDeepak
17.3k11537
$endgroup$
add a comment |
$begingroup$
You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
answered 3 hours ago
DeepakDeepak
17.3k11537
$endgroup$
You are indeed correct. Note that in an indefinite integral, perfectly valid answers can be separated by (any) constant.
The first term in Wolfram's answer can be rewritten:
$3ln{(frac32(x+sqrt{frac49 + x^2}))} = 3ln{(x+sqrt{frac49 + x^2})} + 3lnfrac 32$
and the second term can be rearranged to be identical to your other term.
So your answers are separated by a constant. That's fine. You're right.
answered 3 hours ago
DeepakDeepak
17.3k11537
answered 3 hours ago
DeepakDeepak
17.3k11537
answered 3 hours ago
DeepakDeepak
17.3k11537
answered 3 hours ago
DeepakDeepak
17.3k11537
17.3k11537
add a comment |
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$begingroup$
Have you tried manipulating your answer to look like Wolfram's? Also, note that it's fine if your answer differs from Wolfram's by a constant.
$endgroup$
– Minus One-Twelfth
3 hours ago
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
3 hours ago
1
$begingroup$
One quick way to check the result is to put your formula into Wolfram Alpha (without the constant, which in this case is just adding $3c$ to the result) and subtract the formula Wolfram Alpha gave (again omitting the constant). If the result is a flat constant function then your integral is correct.
$endgroup$
– David K
1 hour ago