Solve the d.e $(x+\sqrt{y^2-xy})dy-ydx=0$ So this is a homogeneous differential equation. 
I use the substitution $y=ux$, $dy=udx+xdu$
And after simplification, got this integral: 
$$
\int\frac{1+\sqrt{u^2-u}}{u\sqrt{u^2-u}}du
$$
I broke it up into two fractions, cancelling out the $\sqrt{u^2-u}$ in one to get, after integration, $\log|u|$ and the other one I have is $$\int \frac{1}{u\sqrt{u^2-u}}du$$
After completing the square I got 
$$
\int\frac{1}{u\sqrt{(u-\frac{1}{2})^2-\frac{1}{4}}}du
$$
This looks like an arcsec trick but I'm not quite sure how to manipulate it to get it into the correct format. 
Any tips? Is this the right approach?  
 A: Put $u=\frac{1}{t}$ in the integral ($\int \frac{du}{u\sqrt{u^2-u}}$)
To get (after simplification)
$$ \int \frac{-dt}{\sqrt{1-t}}$$
Which evaluates to $$2\sqrt{1-t}+C$$
$$=\color{red}{2\sqrt {\frac{u-1}{u}}+C}$$
A: Hint
$$\frac{1 + \sqrt{u^2-u}}{u\cdot\sqrt{u^2 - u}} = \frac{1}{u\cdot \sqrt{u^2-u}} + \frac{1}{u}$$
Once you split into this, you'll have two integrals. The second one is trivial, it will be a log.
The first one can be evaluated by completing the square inside the root as:
$$u \sqrt{\left(u - \frac{1}{2}\right)^2 - \frac{u}{4}}$$
and then make a suitable substitution.
Can you proceed?
Hint 2
At this point, use a suitable substitution like
$$y = u - \frac{1}{2}$$
$$\int\frac{1}{\left(y + \frac{1}{2}\right)\sqrt{y^2 - \frac{1}{4}}}\ \text{d}y$$
Now a cute substitution to take may be
$$y = \frac{\sec(z)}{2} ~~~~~~~~~~~ \text{d}y = \frac{1}{2}\tan(z)\sec(z)\ \text{d}z$$
Can you proceed?
Hint 3
After the substitution, and after having arranged the integral, you'll get
$$\frac{1}{2}\int \frac{2\sec(z)}{\frac{\sec(z)}{2} + \frac{1}{2}}\ \text{d}z$$
simplify the $2$ and then another non standard substitution will help:
$$k = \tan\left(\frac{z}{2}\right) ~~~~~~~~~~~ \text{d}k = \frac{1}{2}\sec^2\left(\frac{z}{2}\right)\ \text{d}z$$
Can you proceed?
A: Take out $u$ again, and then take $1-\frac{1}{u} = t$, then $\frac{du}{u^2} = dt$.
There you go.
