Q69 / 160 · TS EAMCET 2014 · Mathematics

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

  • A
    $3$
  • Correct answer
    B
    $0$
  • C
    $1$
  • D
    $2$

Solution

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 $$ \begin{aligned} & \frac{1}{(1+\sqrt{x})\left(\sqrt{x-x^2}\right)} \\ & =\frac{\sqrt{1-x}\left(A \frac{1}{2 \sqrt{x}}\right)-(A \sqrt{x}+B) \frac{1}{2 \sqrt{2-x}}(-1)}{(\sqrt{1-x})^2} \\ & \Rightarrow \frac{1}{(1+\sqrt{x})\left(\sqrt{x-x^2}\right)}=\frac{(1-x) A+A x+B \sqrt{x}}{(1-x) 2 \sqrt{x} \sqrt{(1-x)}} \\ & \Rightarrow \quad \frac{2(1-x)}{(1+\sqrt{x})}=A+B \sqrt{x} \\ & \Rightarrow \quad 2-2 \sqrt{x}=A+B \sqrt{x} \\ & \Rightarrow \quad A=2, B=2 \\ & \therefore \quad A-B=0 \\ & \end{aligned} $$

Hence, the correct answer is option: 'B'.

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