Evaluate $\int \frac{1}{\sqrt{x^2-1}}dx$ without using trigonometric substitutions

Question

How can I solve the integral

$$\int \frac{1}{\sqrt{x^2-1}}dx$$ without using any trigonometric substitution? I know that $$\frac{d}{dx} \text{arccosh} (x)=±\frac{1}{\sqrt{x^2-1}}$$

Does the fundamental theorem of calculus apply? If so, I could conclude, for $x>0$, the integral as $$\text{arccosh}(x)+C$$

Answer 1

Let $x^2-1=u^2$ thus $dx=\frac{udu}{x}$ but $ x=\sqrt{u^2+1}$ thus integral becomes after simplification $$\int \frac{1}{\sqrt{u^2+1}}du=ln(u+\sqrt{u^2+1^2})+C$$

Answer 2

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 $\frac{1}{2} (t+1-(t-1))$

Answer 3

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.