Question #db6c4

1 Answer
Jul 4, 2017

#int dx/(sqrt(x^2-6)) = ln abs(x +sqrt (x^2-6)) +C#

Explanation:

Note that the function is defined for:

#x in (-oo,-sqrt6) uu (sqrt6,+oo)#

Consider the interval #x in (sqrt6,+oo)# and substitute:

#x =sqrt6cosht#
#dx =sqrt6 sinhtdt#

with #t > 0#

#int dx/(sqrt(x^2-6)) = sqrt6 int (sinhtdt)/(sqrt(6cosh^2t-6)) = int (sinhtdt)/(sqrt(cosh^2t-1) #

Use the hyperbolic identity:

#cosh^2t -1 = sinh^2t#

#int dx/(sqrt(x^2-6)) = int (sinhtdt)/(sqrt(sinh^2t) #

For #t > 0# we have that #sinht > 0#, so:

#int dx/(sqrt(x^2-6)) = int (sinhtdt)/sinht = int dt = t+C #

and undoing the substitution:

#int dx/(sqrt(x^2-6)) = "arcosh"(x/sqrt6) + C = ln (x/sqrt6 +sqrt (x^2/6-1)) +C#

#int dx/(sqrt(x^2-6)) = ln (x +sqrt (x^2-6)) +C#

Similarly in the interval #x in (-oo,-6)# we can substitute:

#x =-sqrt6cosht#
#dx =-sqrt6 sinhtdt#

with #t<0# and conclude that:

#int dx/(sqrt(x^2-6)) = ln abs(x +sqrt (x^2-6)) +C#