Integration with trigonometry substitution

Question

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?

Answer 1

I would Substitute $$t=\sqrt{\frac{4x-3}{x-1}}$$ then we get $$x=\frac{t^2-3}{t^2-4}$$ and $$dx=-\frac{2t}{(t^2-4)^2}dt$$ then we get $$-2\int \frac{t^2}{(t^2-4)^2}dt$$ Can you solve this? for your control we get $$-2(-1/4\, \left( t-2 \right) ^{-1}+1/8\,\ln \left( t-2 \right) -1/4\, \left( t+2 \right) ^{-1}-1/8\,\ln \left( t+2 \right) )+C$$

Answer 2

hint: $$\frac{4x-3}{1-x}=-\frac{4(x-1)+1}{x-1}=-4+\frac{1}{1-x}$$

Answer 3

Using the hint $$\frac{1}{1-x} = 4\sec^2(u)\implies x=\frac{1}{8} (7-\cos (2 u))\implies dx=\frac{1}{4} \sin (2 u)\,du$$ So, after simplifications, $$\int \sqrt{\frac{4x-3}{1-x}}\,dx=\int \sin (u) \cos (u) \sqrt{\tan ^2(u)}\,du$$ I am sure that you can take it from here.