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

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Aicharya · Accepted Answer

Given, $ \int \frac{d x}{(1+\sqrt{x})\left(\sqrt{x^2-x^2}ight)}=\frac{A \sqrt{x}}{\sqrt{1-x}}+\frac{B}{\sqrt{1-x}}+c $ On differentiating both sides, we get $ \begin{aligned} & \frac{1}{(1+\sqrt{x})\left(\sqrt{x-x^2}ight)} \ & =\frac{\sqrt{1-x}\left(A \frac{1}{2 \sqrt{x}}ight)-(A \sqrt{x}+B) \frac{1}{2 \sqrt{2-x}}(-1)}{(\sqrt{1-x})^2} \ & \Rightarrow \frac{1}{(1+\sqrt{x})\left(\sqrt{x-x^2}ight)}=\frac{(1-x) A+A x+B \sqrt{x}}{(1-x) 2 \sqrt{x} \sqrt{(1-x)}} \ & \Rightarrow \quad \frac{2(1-x)}{(1+\sqrt{x})}=A+B \sqrt{x} \ & \Rightarrow \quad 2-2 \sqrt{x}=A+B \sqrt{x} \ & \Rightarrow \quad A=2, B=2 \ & 	herefore \quad A-B=0 \ & \end{aligned} $

If $\int \frac{d x}{(1+\sqrt{x}) \sqrt{x-x^2}}=\frac{A \sqrt{x}}{\sqrt{1-x}}+\frac{B}{\sqrt{1-x}}+C$, where…

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