Finding an integral using a table?
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$begingroup$
Am I correct for pattern matching this integral?
I have
$$int fracsqrt9x^2+4x^2dx$$
Does this pattern match with:
$$int fracsqrta^2 + x^2x^2dx = -fraca^2 + x^2x + ln(x + sqrta^2 + x^2) + c$$
If I factor out the 9, I get
$$= 3 int fracsqrtx^2 + frac49x^2$$
with $a = frac23$
I get:
$$3 left( - fracsqrtfrac49+x^2x + lnleft(x+sqrtfrac49+x^2right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited Mar 4 at 3:44
clathratus
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
$endgroup$
add a comment |
$begingroup$
Am I correct for pattern matching this integral?
I have
$$int fracsqrt9x^2+4x^2dx$$
Does this pattern match with:
$$int fracsqrta^2 + x^2x^2dx = -fraca^2 + x^2x + ln(x + sqrta^2 + x^2) + c$$
If I factor out the 9, I get
$$= 3 int fracsqrtx^2 + frac49x^2$$
with $a = frac23$
I get:
$$3 left( - fracsqrtfrac49+x^2x + lnleft(x+sqrtfrac49+x^2right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited Mar 4 at 3:44
clathratus
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
$endgroup$
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
Mar 4 at 1:58
1
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
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
Mar 4 at 4:18
add a comment |
$begingroup$
Am I correct for pattern matching this integral?
I have
$$int fracsqrt9x^2+4x^2dx$$
Does this pattern match with:
$$int fracsqrta^2 + x^2x^2dx = -fraca^2 + x^2x + ln(x + sqrta^2 + x^2) + c$$
If I factor out the 9, I get
$$= 3 int fracsqrtx^2 + frac49x^2$$
with $a = frac23$
I get:
$$3 left( - fracsqrtfrac49+x^2x + lnleft(x+sqrtfrac49+x^2right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
edited Mar 4 at 3:44
clathratus
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
$endgroup$
Am I correct for pattern matching this integral?
I have
$$int fracsqrt9x^2+4x^2dx$$
Does this pattern match with:
$$int fracsqrta^2 + x^2x^2dx = -fraca^2 + x^2x + ln(x + sqrta^2 + x^2) + c$$
If I factor out the 9, I get
$$= 3 int fracsqrtx^2 + frac49x^2$$
with $a = frac23$
I get:
$$3 left( - fracsqrtfrac49+x^2x + lnleft(x+sqrtfrac49+x^2right) +cright)$$
Is this the right track?
Wolfram winds up with a different answer though:
integration
integration
edited Mar 4 at 3:44
clathratus
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
edited Mar 4 at 3:44
clathratus
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
edited Mar 4 at 3:44
clathratus
5,0361438
edited Mar 4 at 3:44
clathratus
5,0361438
edited Mar 4 at 3:44
clathratus
5,0361438
5,0361438
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
asked Mar 4 at 1:52
Jwan622Jwan622
2,33411632
2,33411632
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
Mar 4 at 1:58
1
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
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
Mar 4 at 4:18
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
Mar 4 at 1:58
1
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
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
Mar 4 at 4:18
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
Mar 4 at 1:58
$begingroup$
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
Mar 4 at 1:58
1
1
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
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
Mar 4 at 4:18
$begingroup$
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
Mar 4 at 4:18
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:
$3ln(frac32(x+sqrtfrac49 + x^2)) = 3ln(x+sqrtfrac49 + x^2) + 3lnfrac 32$
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 Mar 4 at 2:01
DeepakDeepak
17.6k11539
$endgroup$
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:
$3ln(frac32(x+sqrtfrac49 + x^2)) = 3ln(x+sqrtfrac49 + x^2) + 3lnfrac 32$
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 Mar 4 at 2:01
DeepakDeepak
17.6k11539
$endgroup$
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:
$3ln(frac32(x+sqrtfrac49 + x^2)) = 3ln(x+sqrtfrac49 + x^2) + 3lnfrac 32$
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 Mar 4 at 2:01
DeepakDeepak
17.6k11539
$endgroup$
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:
$3ln(frac32(x+sqrtfrac49 + x^2)) = 3ln(x+sqrtfrac49 + x^2) + 3lnfrac 32$
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 Mar 4 at 2:01
DeepakDeepak
17.6k11539
$endgroup$
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:
$3ln(frac32(x+sqrtfrac49 + x^2)) = 3ln(x+sqrtfrac49 + x^2) + 3lnfrac 32$
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 Mar 4 at 2:01
DeepakDeepak
17.6k11539
answered Mar 4 at 2:01
DeepakDeepak
17.6k11539
answered Mar 4 at 2:01
DeepakDeepak
17.6k11539
answered Mar 4 at 2:01
DeepakDeepak
17.6k11539
17.6k11539
add a comment |
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
Mar 4 at 1:58
1
$begingroup$
Actually they are the same, because the answer given by WA is simply yours plus one constant $3 log (3/2)$.
$endgroup$
– xbh
Mar 4 at 2:01
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
Mar 4 at 4:18