The integral int frac{dx}{(1 + sqrt{x}) cdot | vedclass

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Question

(C) Let $I = \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x} \sqrt{1 - x}}$.Substitute $\sqrt{x} = u$,then $dx = 2u \, du$.$I = \int \frac{2u \, du}{(1 + u) u

Vedclass · Accepted Answer

$ - 2\sqrt {\frac{{1 - \sqrt x }}{{1 +\sqrt x }}}  + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(1 + \sqrt{x}) \sqrt{x} \sqrt{1 - x}}$.Substitute $\sqrt{x} = u$,then $dx = 2u \, du$.$I = \int \frac{2u \, du}{(1 + u) u \sqrt{1 - u^2}} = 2 \int \frac{du}{(1 + u) \sqrt{(1 - u)(1 + u)}} = 2 \int \frac{du}{(1 + u) \sqrt{1 - u} \sqrt{1 + u}} = 2 \int \frac{du}{(1 + u)^{3/2} (1 - u)^{1/2}}$.Alternatively,let $1 + \sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$.Since $\sqrt{x} = t - 1$,then $x = (t - 1)^2$,so $1 - x = 1 - (t - 1)^2 = 1 - (t^2 - 2t + 1) = 2t - t^2$.$I = \int \frac{2 dt}{t \sqrt{2t - t^2}} = 2 \int \frac{dt}{t \sqrt{t(2 - t)}} = 2 \int \frac{dt}{t \sqrt{t} \sqrt{2 - t}} = 2 \int \frac{dt}{t^{3/2} \sqrt{2 - t}}$.Let $t = \frac{1}{z}$,then $dt = -\frac{1}{z^2} dz$.$I = 2 \int \frac{-dz/z^2}{(1/z)^{3/2} \sqrt{2 - 1/z}} = 2 \int \frac{-dz/z^2}{z^{-3/2} \sqrt{(2z - 1)/z}} = 2 \int \frac{-dz/z^2}{z^{-3/2} \cdot z^{-1/2} \sqrt{2z - 1}} = 2 \int \frac{-dz}{\sqrt{2z - 1}}$.$I = -2 \int (2z - 1)^{-1/2} dz = -2 \cdot \frac{(2z - 1)^{1/2}}{1/2 \cdot 2} = -2 \sqrt{2z - 1} + c$.Substituting $z = 1/t = 1/(1 + \sqrt{x})$:$I = -2 \sqrt{\frac{2}{1 + \sqrt{x}} - 1} + c = -2 \sqrt{\frac{2 - 1 - \sqrt{x}}{1 + \sqrt{x}}} + c = -2 \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$.

The integral $\int \frac{dx}{(1 + \sqrt{x}) \cdot \sqrt{x} \sqrt{1 - x}}$ is equal to (where $c$ is a constant of integration)

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