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