Any hints on how to compute this integral? Could anyone please give me a hint on how to compute the following integral?
$$\int \sqrt{\frac{x-2}{x^7}} \, \mathrm d x$$
I'm not required to use hyperbolic/ inverse trigonometric functions.

Write $y(x):=\sqrt{\frac {x-2} {x^7}}$. 
Note that $$y'(x)= \frac {7-3x}{x^8} \frac 1 y $$
Hence $$\frac d {dx} x^n y=n x^{n-1} y + x^{n-8} \frac {7-3x}  y.$$
Do the ansatz $$F(x)=\sum_{n=0}^k a_nx^ny ~~~~\text{ and } ~~~~F'(x)=y(x). $$
We get 
$$y\sum_{n=1}^{k}a_n nx^{n-1} +\frac 1 y\sum_{n=0}^{k} a_n x^{n-8} ({7-3x})=y  $$
Thus $$\sum_{n=0}^{k} a_n x^{n-8} ({7-3x})=y^2(1-\sum_{n=1}^{k}a_n nx^{n-1}).$$
Now insert $y^2$
and get 
$$({7-3x})\sum_{n=0}^{k} a_n x^{n-1} = (x-2 )(1-\sum_{n=1}^{k}a_n nx^{n-1}) $$
or equivalently
$$2-x+({7-3x})\sum_{n=0}^{k} a_n x^{n-1}+ (x-2)\sum_{n=1}^{k}a_n nx^{n-1}=0. $$ 

We solve this recursively.  
The lowest order is $x^{-1}$. There we have $7a_0 x^{-1}=0$, so $a_0=0$.  
In constant order: $2+7a_1-3a_0-2a_1=0$, so $a_1=-\frac 2 5$ 
In order $x$: $(-1+7a_2 -3 a_1 +a_1-4 a_2)x=0$ , so $a_2= \frac 1 3 (1+2 
a_1)=\frac 1 {15}$ 
in order $x^2$: $(7a_3-3a_2+2a_2-6 a_3)x^2=0$, so $a_3=a_2=\frac 1 {15}$.  
in order $x^3$: $(7a_4-3a_3+3a_3 - 8a_4)x^4=0$, so $a_4=0$ 
in order $x^n$ for $n>3$: $(7a_{n+1} - 3 a_n +n a_n - 2(n+1) a_{n+1})x^n$, so $a_{n+1}=\frac {n-3}{7-2(n+1)} a_n=0$.

In conclusion it follows that 
$$\int y(x)= const + F(x)= const+ \frac 1 {15} (-6x+ x^2+x^3) y $$
