Question

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

• A. $3$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B

$0$

Explanation for correct option:

Given: $\int \frac{dx}{(1+\sqrt{x})\left(\sqrt{x-x^2}\right)}=\frac{A\sqrt{x}}{\sqrt{1-x}}+\frac{B}{\sqrt{1-x}}+C$

Differentiating both sides,

$\begin{array}{crcll}\Rightarrow & \frac{1}{\left(1+\sqrt{x}\right)\left(\sqrt{x-x^2}\right)}& =& A\frac{\left(\sqrt{1-x}\right){\displaystyle \frac{d}{dx}}{\left[x\right]}^{\frac{1}{2}}-\left(\sqrt{x}\right){\displaystyle \frac{d}{dx}}{\left(1-x\right)}^{{\displaystyle \frac{1}{2}}}}{{\left(\sqrt{1-x}\right)}^2}+B\frac{d}{dx}{\left(1-x\right)}^{-\frac{1}{2}}& \left[\because \frac{d}{\mathrm{dx}}\left[\int f\left(x\right)\mathrm{dx}\right]=f\left(x\right);\frac{d}{\mathrm{dx}}\left(\frac{U}{V}\right)=\frac{V{\displaystyle \frac{\mathrm{dU}}{\mathrm{dx}}}-U{\displaystyle \frac{\mathrm{dV}}{\mathrm{dx}}}}{x^2}\right]\\ \Rightarrow & \frac{1}{\left(1+\sqrt{x}\right)\left(\sqrt{x-x^2}\right)}& =& A\frac{{\displaystyle \left(\left(\sqrt{1-x}\right)\left(\frac{1}{2}\right){\left[x\right]}^{-\frac{1}{2}}\right)-\left(\sqrt{x}\right)\left(\frac{1}{2}\right)\left(-1\right){\left(1-x\right)}^{-\frac{1}{2}}}}{{\left(\sqrt{1-x}\right)}^2}+B\left(-\frac{1}{2}\right)\left(-1\right){\left(1-x\right)}^{-\frac{3}{2}}& \left[\because \frac{d}{\mathrm{dx}}f\left[g\left(x\right)\right]=f\text{'}\left[g\left(x\right)\right]·g\text{'}\left(x\right)\right]\\ \Rightarrow & \frac{1}{\left(\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(\sqrt{1-x}\right)}& =& \frac{1}{2}A\frac{{\displaystyle \left(\frac{\left(\sqrt{1-x}\right)}{\sqrt{x}}\right)+\left(\frac{\sqrt{x}}{\sqrt{1-x}}\right)}}{{\left(\sqrt{1-x}\right)}^2}+B\left(\frac{1}{2}\right)\frac{1}{\sqrt{{\left(1-x\right)}^3}}& \\ \Rightarrow & \frac{2}{\left(\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(\sqrt{1-x}\right)}& =& A\frac{{\displaystyle 1-x+x}}{\left(\sqrt{x}\right)\left(\sqrt{{\left(1-x\right)}^3}\right)}+B\frac{1}{\sqrt{{\left(1-x\right)}^3}}& \\ \Rightarrow & \frac{2}{\left(\sqrt{x}\right)\left(1+\sqrt{x}\right)\left(\sqrt{1-x}\right)}& =& \frac{{\displaystyle A+B\sqrt{x}}}{\left(\sqrt{1-x}\right)\left(\sqrt{x}\right)\left(1-x\right)}& \\ \Rightarrow & \frac{2}{1+\sqrt{x}}& =& \frac{A+B\sqrt{x}}{1-x}& \\ \Rightarrow & A+B\sqrt{x}& =& \frac{2(1-x)}{1+\sqrt{x}}& \\ \Rightarrow & A+B\sqrt{x}& =& \frac{2(1-x)}{1+\sqrt{x}}\times \frac{1-\sqrt{x}}{1-\sqrt{x}}& \\ \Rightarrow & A+B\sqrt{x}& =& \frac{2(1-x)\left(1-\sqrt{x}\right)}{1-{\left(\sqrt{x}\right)}^2}& \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ \Rightarrow & A+B\sqrt{x}& =& 2-2\sqrt{x}& \end{array}$

Comparing both sides, we get $A=2,B=-2$

$\therefore A+B=2+(-2)=0$

Hence, option(B) is correct.