I want to calculate the answer of the integral $$\int\frac{dx}{x\sqrt{x^2-1}}$$ I use the substitution $x=\cosh(t)$ ($t \ge 0$) which yields $dx=\sinh(t)\,dt$. By using the fact that $\cosh^2(t)-\sinh^2(t)=1$ we can write $x^2-1=\cosh^2(t)-1=\sinh^2(t)$. Since $t\ge 0$, $\sinh(t)\ge 0$, and we have $\sqrt{x^2-1}=\sinh(t)$. Now, by substituting for $x$, $dx$, and $\sqrt{x^2-1}$ in the first integral, we have $$\int\frac{dt}{\cosh(t)}$$ Since $\cosh(t)=(e^t+e^{-t})/2$ by substituting this in this integral we have $$2\int\frac{dt}{e^t+e^{-t}}.$$ Now by multiplying numerator and denominator in $e^t$ one can write:$$2\int\frac{e^t\,dt}{1+e^{2t}}$$ Now by using $z=e^t$ in this integral one can write ($dz=e^t\,dt$): $$2\int\frac{dz}{1+z^2}$$ So we have $$\int\frac{dt}{\cosh(t)}=2\arctan(z)=2\arctan(e^t)$$ On the other hand $$t=\cosh^{-1}(x)=\ln\left(x+\sqrt{x^2-1}\right)$$ So we have $$\int\frac{dx}{x\sqrt{x^2-1}}=\int\frac{dt}{\cosh(t)}=2\arctan(\exp(\ln(x+\sqrt{x^2-1})))$$ which yields
$$\int\frac{dx}{x\sqrt{x^2-1}}=2\arctan(x+\sqrt{x^2-1})$$
but this answer is wrong. The true answer can be obtained by direct substitution $u=\sqrt{x^2-1}$ and is
$$\int\frac{dx}{x\sqrt{x^2-1}}=\arctan(\sqrt{x^2-1})$$
I don't want to know the answer of the integral. I want to know what I did wrong? Can somebody help?
