Finding an integral using a table?
$begingroup$
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
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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 |
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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 |
add a comment |
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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