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

1 Answer
Mar 4, 2018

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