What is the value of the following integral ?!
$\int\sqrt{(1+(\frac{x}{h})^{2})^{-1}-b}~~~dx$
I have tried on it as : Let $y=1+(\frac{x}{h})^{2}$ $dy=\frac{2x}{h^{2}}dx$ Thus $dx=\frac{h^{2}}{2x}dy$ and $\frac{h}{x}=\frac{1}{\sqrt{y-1}}$ Hence the integral become $\int\sqrt{(y)^{-1}-b} \frac{h^{2}}{2x}~~dy=\frac{h}{2}\int\sqrt{y^{-1}-b} \frac{1}{\sqrt{y-1}}=\frac{h}{2}\int\sqrt{\frac{y^{-1}-b}{y-1}}~~dy=\frac{h}{2}\int\sqrt{\frac{1-yb}{y(y-1)}}~~dy$ But i can not complete it , can you help me ??