$$ \int {\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}^{1/2}dx $$
Any suggestion on how to solve it is appreciated.
$$ \int {\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}^{1/2}dx $$
Any suggestion on how to solve it is appreciated.
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HINT:
Let $\sqrt x=\cos2t\implies0\le2t\le\dfrac\pi2$
$$\dfrac{1-\sqrt x}{1+\sqrt x}=\tan^2t\implies\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}=+\tan t$$
$$x=\cos^22t\implies dx=-2\cos2t\sin2t\ dt$$
$$ \int {\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)}^{1/2}dx=-4\int\cos2t\sin^2t\ dt$$
$$=2\int\cos2t(\cos2t-1)dt$$
$$=\int(1+\cos4t-2\cos2t)dt$$
Can you take it from here?