How do you find the integral #intsqrt(x^2-1)/xdx# ?

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Answer 1 · 2018-03-04T08:54:35

The answer is #=sqrt(x^2-1)-arctan(sqrt(x^2-1))+C#

We need

#int(dx)/(x^2+1)=arctanx+C#

Perform this integral by substitution

Let #u=sqrt(x^2-1)#, #=>#, #x^2=u^2+1#

#du=(2xdx)/(2sqrt(x^2-1))=(xdx)/(sqrt(x^2-1))#

Therefore,

#int(sqrt(x^2-1)dx)/(x)=intsqrt(x^2-1)/x*sqrt(x^2-1)/x*dx#

#=int((x^2-1)du)/(x^2)#

#=int(u^2)/(u^2+1)*du#

#=int(u^2+1-1)/(u^2+1)du#

#=intdu-int(du)/(u^2+1)#

#=u+arctan(u)#

#=sqrt(x^2-1)-arctan(sqrt(x^2-1))+C#