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

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Answer 1 · 2017-06-23T03:55:09

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

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

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 $theta=sec^-1(x^2/2)$ :

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

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