How can we calculate $\int \:\frac{1}{\sqrt{x^2-1}}dx$

Question

Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$

I attempted to solve it intergradation by parts by doing a

$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$

and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:

but as I start doing, it gets more complicated. What is the right direction to solve this?

Answer 1

For $x^2-1$ in the radical use $x=\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$

For $x^2+1$ in the radical use $x=\tan\theta$ for the same reason

For $1-x^2$ in the radical use $x=\sin\theta$ as $1-\sin^2\theta=\cos^2\theta$

Answer 2

the solution is $= \operatorname{Argch(x)}$

because $\operatorname{Argch}^{\prime}(x)= \frac{1}{\sqrt{x^2-1}}$.

Answer 3

Here's an alternative way doing it. Rewrite the integral as:

$\int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}} dx = \int \frac{1 + \frac{x}{\sqrt{x^2-1}} dx}{x + \sqrt{x^2-1}}$.

Now use a simple substitution $u=x+\sqrt{x^2-1}$, so that the integral becomes:

$\int \frac{du}{u} = \log u + C = \log{x+\sqrt{x^2-1}}+C.$

Answer 4

HINT:

Substitute $\displaystyle x=\frac1y$ and then $\displaystyle 1-y^2=u^2$

or directly, $\displaystyle x=\frac1y=\frac1{\sqrt{1-u^2}}\implies x^2-1=\frac{u^2}{1-u^2}$