How do you find the integral of #int 1/(xsqrt(x^4-4)#?

1 Answer
mason m
Jun 23, 2017

#intdx/(xsqrt(x^4-4))=1/4sec^-1(x^2/2)+C#

Explanation:

#I=intdx/(xsqrt(x^4-4))#

Try the substitution #x^2=2sectheta#. This implies that #2xdx=2secthetatanthetad theta#.

It also implies that #sqrt(x^4-4)=sqrt(4sec^2theta-4)=2sqrt(sec^2theta-1)=2tantheta#. Then:

#I=1/2int(2xdx)/(x^2sqrt(x^4-4))#

#I=1/2int(2secthetatanthetad theta)/(2sectheta(2tantheta))#

#I=1/4intd theta#

#I=1/4theta#

From #x^2=2sectheta# we see that #theta=sec^-1(x^2/2)#:

#I=1/4sec^-1(x^2/2)+C#

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