Think about substitution

We attempt the substitution $u = \sqrt{x^n - 1}$ . Note that we obtain $du = \dfrac{nx^{n-1}}{2\sqrt{x^n-1}} dx \Rightarrow dx = \dfrac{2\sqrt{x^n-1}}{nx^{n-1}} du$ . This yields:

$\int^\infty_1 \frac{n}{x\sqrt{x^n-1}} dx = \int^{\infty}_{0} \frac{2}{x^n} du = 2 \int^\infty_0 \frac{1}{1+u^2} du = 2\: \text{arctan}(u) \bigg|^\infty_0 = 2\left(\frac{\pi}{2} - 0\right) = \pi$

So the answer is $\cos(\pi) = \boxed{-1}$ .

Simply substitute $\displaystyle x^{\frac{n}{2}}=\sec\theta$ and after simplifying, we need to solve $\int_0^{\frac{\pi}{2}}2\ d\theta$ and thus, we get $\alpha=\pi$ making the answer $-1$

Substitute $u=x^n-1$ , then we obtain $du=nx^{n-1}dx$ and $\displaystyle dx=\frac{1}{nx^{n-1}}du$ . Now we get:

$\displaystyle \int _{ 0 }^{ \infty }{ \frac { n }{ x\sqrt { u } \cdot n{ x }^{ n-1 } } du } =\int _{ 0 }^{ \infty }{ \frac { 1 }{ { x }^{ n }\sqrt { u } } du } =\int _{ 0 }^{ \infty }{ \frac { 1 }{ \left( u+1 \right) \sqrt { u } } du }$

Substitute again $u=\tan ^2 \theta$ , then we obtain $du = 2\tan \theta \sec ^2 \theta d \theta$ . Apply this and we get:

$\displaystyle\int _{ 0 }^{ \frac{\pi}{2} }{ \frac { 2\tan \theta \sec ^2 \theta }{ \left( \tan ^2 \theta +1 \right) \tan \theta } d\theta } =\int _{ 0 }^{ \frac{\pi}{2} }{2d\theta}=\pi$

Thus the answer is $\cos\pi = \boxed{-1}$