How do you integrate:
$$\int \sqrt{\frac{(4x-3)}{1-x}}dx$$
hint given was $\frac{1}{1-x} = 4\sec^2(u)$ do i need to use trigonometric substitution for this? Even so, not sure how to solve it
After trying, the answer i got was $$u/2 +{1/2\sin u\cos u} + c$$ And after substitution to get back x, i got $$ \frac{\arccos(2\sqrt(1-x))}{2} + \sqrt(1-x)\sqrt(4x-3) + c$$
Is this correct?