Integral of $\frac{1}{\sqrt{x^2-1}}$ w/o trig. subst.? How can I solve the integral 
$$\int \frac{1}{\sqrt{x^2-1}}$$ without using trigonometric substitution? Let $x>0$.
I know that $$\frac{d}{dx} arcosh(x)=±\frac{1}{\sqrt{x^2-1}}$$
So does fundamental theorem of calculus apply and I could conclude the integral as $$arcosh(x)+C$$?
 A: I don't see why you won't do a trigonometric sub, (put $x=\sec \theta$) , but anyway, this integral can be written as:
$$\int \frac{dx}{x\sqrt{1-\frac{1}{x^2}}}$$ 
Substitute $1-\frac{1}{x^2}=t^2$, 
You get $$\frac{2}{x^3}dx=2tdt$$
$$\implies \frac{dx}{x}.\frac{1}{x^2}=tdt$$
$$\implies \frac{dx}{x}.(t^2-1)=-tdt$$
$$\implies \frac{dx}{x}=\frac{-tdt}{t^2-1}$$
Now, put everything back in the integral to get:
$$\int \frac{-dt}{t^2-1}$$
Which may be done by partial fractions or by splitting the numerator as $\frac{1}{2} (t+1-(t-1))$
A: Let $x^2-1=u^2$ thus $dx=\frac{udu}{x}$ but $ x=\sqrt{u^2+1}$ thus integral becomes after simplification $$\int \frac{1}{\sqrt{u^2+1}}du=ln(u+\sqrt{u^2+1^2})+C$$
A: The function $\operatorname{arcosh}$ is only defined over $[1,\infty)$ where it satisfies $x=\cosh\operatorname{arcosh}x$, so
$$
1=\sinh\operatorname{arcosh}x\operatorname{arcosh}'x
$$
which means
$$
\operatorname{arcosh}'x=\frac{1}{\sin\operatorname{arcosh}x}=
\frac{1}{\sqrt{x^2-1}}
$$
for $x>1$ (it is not differentiable at $1$).
Thus the most general antiderivative of your function is
$$
\begin{cases}
\operatorname{arcosh}x+c_+ & \text{for $x>1$} \\[6px]
-\operatorname{arcosh}(-x)+c_- & \text{for $x<-1$}
\end{cases}
$$
where $c_+$ and $c_-$ are arbitrary constants.
There's no need to do substitutions, trigonometric or not, if you already know about that function.
