How do you evaluate the integral $int dx/(x^2sqrt(x^2-3))$ ?

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Answer 1 · 2017-08-31T17:51:17

After using $x=sqrt(3)*secu$ and $dx=sqrt(3)*secu*tanu*du$ transforms, this integral became,

$int (sqrt(3)*secu*tanu*(du))/[3(secu)^2*sqrt(3)*tanu]$

= $1/3*int (du)/secu$

= $1/3*int cosu*du$

= $1/3*sinu+C$

After using $x=sqrt(3)*secu$ , $secu=x/sqrt(3)$ , $tanu=sqrt(x^2-3)/sqrt(3)$ and $sinu=tanu/secu=sqrt(x^2-3)/x$ inverse transforms, I found,

$int (dx)/[x^2*sqrt(x^2-3)]=sqrt(x^2-3)/(3x)+C$

I used $x=sqrt(3)*secu$ and $dx=sqrt(3)*secu*tanu*du$ transform

How do you evaluate the integral int dx/(x^2sqrt(x^2-3))int dx/(x^2sqrt(x^2-3))?