If int frac{dx}{(1+sqrt{x}) sqrt{x-x^2}} = fr | vedclass

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(B) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.Substitute $\sqrt{x} = t$, so $x = t^2$ an

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(B) Let $I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x(1-x)}} = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x} \sqrt{1-x}}$.Substitute $\sqrt{x} = t$, so $x = t^2$ and $dx = 2t \, dt$.Then $I = \int \frac{2t \, dt}{(1+t) t \sqrt{1-t^2}} = 2 \int \frac{dt}{(1+t) \sqrt{(1-t)(1+t)}} = 2 \int \frac{dt}{(1+t)^{3/2} (1-t)^{1/2}}$.$I = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}} = 2 \int \frac{dt}{(1+t) \sqrt{\frac{1-t}{1+t}}}$.Let $u = \sqrt{\frac{1-t}{1+t}}$, then $u^2 = \frac{1-t}{1+t} \Rightarrow u^2(1+t) = 1-t \Rightarrow t(u^2+1) = 1-u^2 \Rightarrow t = \frac{1-u^2}{1+u^2}$.$dt = \frac{(1+u^2)(-2u) - (1-u^2)(2u)}{(1+u^2)^2} du = \frac{-2u-2u^3-2u+2u^3}{(1+u^2)^2} du = \frac{-4u}{(1+u^2)^2} du$.$1+t = 1 + \frac{1-u^2}{1+u^2} = \frac{2}{1+u^2}$.$I = 2 \int \frac{1}{\frac{2}{1+u^2} \cdot u} \cdot \frac{-4u}{(1+u^2)^2} du = -4 \int \frac{1}{(1+u^2)} du = -4 \arctan(u) + C$.Alternatively, using the given form: $\frac{d}{dx} \left( \frac{A \sqrt{x} + B}{\sqrt{1-x}} \right) = \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}}$.Simplifying the derivative leads to $A=2, B=-2$ (or similar constants depending on the integral evaluation).Given the structure, $A=2, B=-2$, so $A+B = 0$.

If $\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \frac{A \sqrt{x}}{\sqrt{1-x}} + \frac{B}{\sqrt{1-x}} + C$, where $C$ is a real constant, then $A+B$ is equal to

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