Finding an integral using a table?












4












$begingroup$


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:



enter image description here






integration







share|cite|improve this question











$endgroup$








  • 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
















4












$begingroup$


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:



enter image description here






integration







share|cite|improve this question











$endgroup$








  • 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














4












4








4


2



$begingroup$


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:



enter image description here






integration







share|cite|improve this question











$endgroup$




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:



enter image description here






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 4 at 3:44









clathratus

4,8901338




4,8901338










asked Mar 4 at 1:52









Jwan622Jwan622

2,24111632




2,24111632








  • 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




    $begingroup$
    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










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.






share|cite|improve this answer









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    6












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 4 at 2:01









        DeepakDeepak

        17.4k11539




        17.4k11539






























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