# How to integrate $\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$?

How to integrate $$\int_1^\infty \frac{dx}{x^2\sqrt{x^2-1}}$$

I tried both $t=\sqrt{x^2-1}$ and $t=\sin x$ but didn’t reach the right result.

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\begin{align}&\overbrace{\color{#66f}{\large%
\int_{1}^{\infty}{\dd x \over x^{2}\root{x^{2} – 1}}}}
^{\ds{\dsc{x} = \dsc{1 \over t}\ \imp\ \dsc{t} = \dsc{1 \over x}}}\ =\
\int_{1}^{0}{-\,\dd t/t^{2} \over t^{-2}\root{1/t^{2} – 1}}
=\int_{0}^{1}{t\,\dd t \over \root{1 – t^{2}}}
=\left. -\root{1 – t^{2}}\right\vert_{0}^{1}=\color{#66f}{\LARGE 1}
\end{align}

Hint: Let $x \mapsto 1/y$, then we have $\mathrm{d}x = -\mathrm{d}y/y^2$ which implies that $-\mathrm{d}y = \mathrm{d}x/x^2$. Hence
$$\int_1^\infty \frac1{\sqrt{x^2-1\,}\,}\frac{\mathrm{d}x}{x^2} = \int_1^0 \frac{-\mathrm{d}y}{\sqrt{(1/y)^2-1\,}\,} = \int_0^1 \frac{y\,\mathrm{d}y}{\sqrt{1-y^2\,}\,}$$
From here a simple and (obvious) substitution will complete the work.

Here’s the easiest way:

$$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \frac{dx}{x^3 \sqrt{1 – \frac{1}{x^2}}}$$

Now substitute $u = 1 – \frac{1}{x^2}$

$$x=\sec u, \;\mathrm{d}x=\tan u\sec u\,\mathrm{d}u$$

If you decide to $x\mapsto \frac1x$ then substitute $y=1-x^2,\;\mathrm{d}y=-2x\,\mathrm{d}x$

\begin{align} \int_1^\infty\frac{\mathrm{d}x}{x^2\sqrt{x^2-1}} &=\frac12\int_1^\infty\frac{\mathrm{d}x}{x^{3/2}\sqrt{x-1}}\\ &=\frac12\int_0^\infty\frac{x^{-1/2}\mathrm{d}x}{(1+x)^{3/2}}\\ &=\frac12\mathrm{B}\left(\frac12,1\right)\\ &=\frac12\frac{\Gamma\left(\frac12\right)\Gamma(1)}{\Gamma\left(\frac32\right)}\\ &=\frac12\frac{\Gamma\left(\frac12\right)\Gamma(1)}{\frac12\Gamma\left(\frac12\right)}\\[9pt] &=1 \end{align}

we set $x=\cosh(t)$ then we have
$$dx=sinh(t)dt$$
thus we get $$\int\frac{\sinh(t)}{\cosh(t)^2\sqrt{\cosh(t)^2-1}}\,dt$$

I recommend regular trig-sub, you should get a pretty simple integral at the end. Set up a right triangle with some angle $\theta$, hypotenuse length $x$ and the side opposite $\theta$ with length $1$. You should get $$\frac{1}{x^2} = \sin^2(\theta), \quad \tan(\theta) = \frac{1}{\sqrt{x^2-1}}, \quad dx = -\csc(\theta)\cot(\theta)d\theta$$ Hence $$\int \frac{dx}{x^2\sqrt{x^2-1}} = \int \ -sin^2(\theta)\tan(\theta)\csc(\theta)\cot(\theta)d\theta \\ = \int -\sin(\theta)d\theta$$

Apply the substitution $x=sec(u)$, then $dx=\frac{\tan \left(u\right)}{\cos \left(u\right)}du$.

$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{\tan \left(u\right)}{\cos \left(u\right)}du$$

$$\int \frac{1}{\sec ^2\left(u\right)\sqrt{\sec ^2\left(u\right)-1}}\frac{1}{\cos^2 \left(u\right)}du$$

Remember that $\frac {1}{\cos(u)}=\sec(u)$ :

$$\int \frac{\sin \left(u\right)}{\sqrt{\sec ^2\left(u\right)-1}}du$$

And $\sec ^2\left(x\right)=1+\tan ^2\left(x\right)$:

$$\int \frac{\sin \left(u\right)}{\sqrt{-1+1+\tan ^2\left(u\right)}}du$$

Then:

$$\int \frac{\sin \left(u\right)}{\sqrt{\tan ^2\left(u\right)}}du$$

$$if \int \frac{g\left(x\right)}{f\left(x\right)}=F\left(x\right)\mathrm{,\:then\:}\:\int \frac{g\left(x\right)}{\sqrt{f^2\left(x\right)}}=\frac{f\left(x\right)}{\sqrt{f^2\left(x\right)}}F\left(x\right)$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \frac{\sin \left(u\right)}{\tan \left(u\right)}du$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\int \cos \left(u\right)du$$

$$\frac{\tan \left(u\right)}{\sqrt{\left(\tan \left(u\right)\right)^2}}\sin \left(u\right)$$

Substitute back, $u=arcsec(u)$:

$$\frac{\tan \left(arcsec \left(x\right)\right)}{\sqrt{\left(\tan \left(arcsec \left(x\right)\right)\right)^2}}\sin \left(arcsec \left(x\right)\right)$$

Simplifies to:

$$\sqrt{1-\frac{1}{x^2}}+C$$

Now compute the boundaries and you will find $1-0=1$.