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