Integrate $\int \frac{dx}{\sqrt {7-6x}-x^2}$ $$\int \frac{dx}{\sqrt {7-6x}-x^2}$$
Have no idea how to start with this. How do i make a perfect square of the variable?

I assume you are trying to solve the integral
$$I:=\int \frac{\mathrm d x}{\sqrt{7-6x-x^2}}$$
because you mentioned to make a perfect square.
Since $$7-6x-x^2= 16-(x+3)^2= \frac{1}{16}\left(1- \left( \frac{x+3}{4}\right)^2\right)$$
the integral becomes

$$I= \int \frac{\mathrm d x}{\frac{1}{4}\sqrt{1- \left( \frac{x+3}{4}\right)^2 }} = \arcsin \left( \frac{x+3}{4}\right).$$ 

In the last step you need to substitute $t= \frac{x+3}{2}$.
