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.
 A: Write your integrand in the form $$\frac{\sqrt{x-2}}{x^{7/2}}$$ and then substitute $$u=\sqrt{x}$$ so you will get $$2\int\frac{\sqrt{u^2-2}}{u^6}du$$ after this substitute $$u=\sqrt{2}\sec(s)$$ to get $$2\sqrt{2}\int\frac{\sin^2(s)\cos^2(s)}{4\sqrt{2}}ds$$
A: 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 
$$\int y(x)= const + F(x)= const+ \frac 1 {15} (-6x+ x^2+x^3) y $$
A: With $y:=\dfrac1x$,
$$\int\sqrt{\frac{x-2}{x^7}}dx=-\int\sqrt{\left(\frac1y-2\right)y^7}\,\frac{dy}{y^2}=-\int y\sqrt{1-2y}\,dy.$$
Then by parts,
$$-\int y\sqrt{1-2y}\,dy=\frac13y(1-2y)^{3/2}-\frac13\int(1-2y)^{3/2}dy=\frac13y(1-2y)^{3/2}+\frac1{15}(1-2y)^{5/2}$$
$$=\frac1{3x}\left(1-\frac2x\right)^{3/2}+\frac1{15}\left(1-\frac2x\right)^{5/2}.$$
A: Try the substitution
$$
u=\frac{x-2}{x}
$$ 
or equivalently
$$
x=\frac{2}{1-u}
$$
