We need
int(dx)/(x^2+1)=arctanx+C∫dxx2+1=arctanx+C
Perform this integral by substitution
Let u=sqrt(x^2-1)u=√x2−1, =>⇒, x^2=u^2+1x2=u2+1
du=(2xdx)/(2sqrt(x^2-1))=(xdx)/(sqrt(x^2-1))du=2xdx2√x2−1=xdx√x2−1
Therefore,
int(sqrt(x^2-1)dx)/(x)=intsqrt(x^2-1)/x*sqrt(x^2-1)/x*dx∫√x2−1dxx=∫√x2−1x⋅√x2−1x⋅dx
=int((x^2-1)du)/(x^2)=∫(x2−1)dux2
=int(u^2)/(u^2+1)*du=∫u2u2+1⋅du
=int(u^2+1-1)/(u^2+1)du=∫u2+1−1u2+1du
=intdu-int(du)/(u^2+1)=∫du−∫duu2+1
=u+arctan(u)=u+arctan(u)
=sqrt(x^2-1)-arctan(sqrt(x^2-1))+C=√x2−1−arctan(√x2−1)+C
