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

1 Answer
Mar 4, 2018

Answer:

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

Explanation:

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#