Integrate $frac3sqrt-5x^2-4sqrt5x+2$

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Integrate $frac3sqrt-5x^2-4sqrt5x+2$




For this question I first factored out the 3, seeing as its a constant



$$3intfrac1sqrt-5x^2-4sqrt5x+2dx$$



I then noticed that this integral has a similar form to the integral of $arcsin(x)$



$$intfrac1sqrta^2-x^2dx=arcsin(fracxa)+C, vert x vert lt a$$



The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4sqrt5x+2$$ to $$a^2-x^2$$



Any help or ideas would be highly appreciated!







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  • $$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
    – Hazem Orabi
    Aug 29 at 10:19















up vote
3
down vote

favorite
1













Integrate $frac3sqrt-5x^2-4sqrt5x+2$




For this question I first factored out the 3, seeing as its a constant



$$3intfrac1sqrt-5x^2-4sqrt5x+2dx$$



I then noticed that this integral has a similar form to the integral of $arcsin(x)$



$$intfrac1sqrta^2-x^2dx=arcsin(fracxa)+C, vert x vert lt a$$



The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4sqrt5x+2$$ to $$a^2-x^2$$



Any help or ideas would be highly appreciated!







share|cite|improve this question






















  • $$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
    – Hazem Orabi
    Aug 29 at 10:19













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1






Integrate $frac3sqrt-5x^2-4sqrt5x+2$




For this question I first factored out the 3, seeing as its a constant



$$3intfrac1sqrt-5x^2-4sqrt5x+2dx$$



I then noticed that this integral has a similar form to the integral of $arcsin(x)$



$$intfrac1sqrta^2-x^2dx=arcsin(fracxa)+C, vert x vert lt a$$



The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4sqrt5x+2$$ to $$a^2-x^2$$



Any help or ideas would be highly appreciated!







share|cite|improve this question















Integrate $frac3sqrt-5x^2-4sqrt5x+2$




For this question I first factored out the 3, seeing as its a constant



$$3intfrac1sqrt-5x^2-4sqrt5x+2dx$$



I then noticed that this integral has a similar form to the integral of $arcsin(x)$



$$intfrac1sqrta^2-x^2dx=arcsin(fracxa)+C, vert x vert lt a$$



The similarities between these two integrals is clear, but I dont know how to get from $$-5x^2-4sqrt5x+2$$ to $$a^2-x^2$$



Any help or ideas would be highly appreciated!









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 29 at 15:56









tarit goswami

1,137219




1,137219










asked Aug 29 at 10:09









Pablo

33612




33612











  • $$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
    – Hazem Orabi
    Aug 29 at 10:19

















  • $$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
    – Hazem Orabi
    Aug 29 at 10:19
















$$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
– Hazem Orabi
Aug 29 at 10:19





$$ -5x^2-4sqrt5xcolorred-4colorred+4+2 = 6-left(5x^2+4sqrt5x+4right) $$
– Hazem Orabi
Aug 29 at 10:19











2 Answers
2






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A rather messy completing the square:



$-5x^2-4sqrt5x+2$



$-5left(x^2+frac4sqrt5x5right)+2$



$2-5left(x+frac2sqrt55 right)^2+4$



$(sqrt6)^2-left(sqrt5left(x+frac2sqrt55 right)right)^2$






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  • that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
    – Pablo
    Aug 29 at 12:44

















up vote
0
down vote













Hint: Write
$$-5x^2-4sqrt5x+2=6-(sqrt5x+2)^2$$
and let substitution $sqrt5x+2=sqrt6sintheta$.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    A rather messy completing the square:



    $-5x^2-4sqrt5x+2$



    $-5left(x^2+frac4sqrt5x5right)+2$



    $2-5left(x+frac2sqrt55 right)^2+4$



    $(sqrt6)^2-left(sqrt5left(x+frac2sqrt55 right)right)^2$






    share|cite|improve this answer






















    • that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
      – Pablo
      Aug 29 at 12:44














    up vote
    1
    down vote



    accepted










    A rather messy completing the square:



    $-5x^2-4sqrt5x+2$



    $-5left(x^2+frac4sqrt5x5right)+2$



    $2-5left(x+frac2sqrt55 right)^2+4$



    $(sqrt6)^2-left(sqrt5left(x+frac2sqrt55 right)right)^2$






    share|cite|improve this answer






















    • that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
      – Pablo
      Aug 29 at 12:44












    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    A rather messy completing the square:



    $-5x^2-4sqrt5x+2$



    $-5left(x^2+frac4sqrt5x5right)+2$



    $2-5left(x+frac2sqrt55 right)^2+4$



    $(sqrt6)^2-left(sqrt5left(x+frac2sqrt55 right)right)^2$






    share|cite|improve this answer














    A rather messy completing the square:



    $-5x^2-4sqrt5x+2$



    $-5left(x^2+frac4sqrt5x5right)+2$



    $2-5left(x+frac2sqrt55 right)^2+4$



    $(sqrt6)^2-left(sqrt5left(x+frac2sqrt55 right)right)^2$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 29 at 10:25

























    answered Aug 29 at 10:19









    Bruce

    526113




    526113











    • that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
      – Pablo
      Aug 29 at 12:44
















    • that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
      – Pablo
      Aug 29 at 12:44















    that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
    – Pablo
    Aug 29 at 12:44




    that is an incredibly messy completing teh square, youre right. any "easier" methods? or is this the only one?
    – Pablo
    Aug 29 at 12:44










    up vote
    0
    down vote













    Hint: Write
    $$-5x^2-4sqrt5x+2=6-(sqrt5x+2)^2$$
    and let substitution $sqrt5x+2=sqrt6sintheta$.






    share|cite|improve this answer
























      up vote
      0
      down vote













      Hint: Write
      $$-5x^2-4sqrt5x+2=6-(sqrt5x+2)^2$$
      and let substitution $sqrt5x+2=sqrt6sintheta$.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Hint: Write
        $$-5x^2-4sqrt5x+2=6-(sqrt5x+2)^2$$
        and let substitution $sqrt5x+2=sqrt6sintheta$.






        share|cite|improve this answer












        Hint: Write
        $$-5x^2-4sqrt5x+2=6-(sqrt5x+2)^2$$
        and let substitution $sqrt5x+2=sqrt6sintheta$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 29 at 10:15









        Nosrati

        22k51747




        22k51747



























             

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