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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?

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  • $\begingroup$ Use $u$ substitution and write the square root as two square roots. So if $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $\endgroup$ – Latin Wolf Mar 29 '18 at 13:48
  • $\begingroup$ @ Latin Wolf, mind if you could elaborate more on using u substitution? $\endgroup$ – Isabella.T Mar 29 '18 at 14:02
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    $\begingroup$ if you use the hint given in the problem you should end up with $\int \sin ^2udu$ $\endgroup$ – Lozenges Mar 29 '18 at 14:06
  • $\begingroup$ @ Lozenges, how do i use the hint for the numerator? $\endgroup$ – Isabella.T Mar 29 '18 at 14:08
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    $\begingroup$ from the hint $4-4x =\cos ^2u $ we get $4x-3=\sin ^2u$ and $4 \text{dx} = 2 \sin u \cos u \text{du}$ $\endgroup$ – Lozenges Mar 29 '18 at 14:11
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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 $$-2(-1/4\, \left( t-2 \right) ^{-1}+1/8\,\ln \left( t-2 \right) -1/4\, \left( t+2 \right) ^{-1}-1/8\,\ln \left( t+2 \right) )+C$$

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  • $\begingroup$ @ Dr. Sonnhard Graubner if I am not wrong $\int \frac{t^2}{(t^2-4)^2} dt= \int \frac{1}{t^2-4} dt + 4\int \frac{1}{(t^2-4)^2} dt$ $\endgroup$ – Latin Wolf Mar 29 '18 at 13:58
  • $\begingroup$ @ Dr. Sonnhard Graubner, I am still unable to get the solution, i tried using trigo substitution on your last equation. $\endgroup$ – Isabella.T Mar 29 '18 at 14:00
  • $\begingroup$ use that $$\frac{t^2}{(t^2-4)^2}=1/4\, \left( t-2 \right) ^{-2}-1/8\, \left( t+2 \right) ^{-1}+1/8\, \left( t-2 \right) ^{-1}+1/4\, \left( t+2 \right) ^{-2} $$ $\endgroup$ – Dr. Sonnhard Graubner Mar 29 '18 at 14:02
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hint: $$\frac{4x-3}{1-x}=-\frac{4(x-1)+1}{x-1}=-4+\frac{1}{1-x}$$

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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.

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  • $\begingroup$ Isn't your last integrand equal to $\sin^2(u)$ $\endgroup$ – Latin Wolf Mar 30 '18 at 3:04
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    $\begingroup$ @LatinWolf. For sure ! I did let the OP the task of finishing the work. $\endgroup$ – Claude Leibovici Mar 30 '18 at 3:23

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