Question

$\int \frac{dx}{(1+\sqrt{x})\sqrt{(x-x^2)}}$ is equal to

• A. $\frac{1+\sqrt{x}}{(1-x{)}^2}+c$
• B. $\frac{1+\sqrt{x}}{(1+x{)}^2}+c$
• C. $\frac{1-\sqrt{x}}{(1-x{)}^2}+c$
• D. $\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+c$

Answer 1

The correct option is D $\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+c$

$I=\int \frac{dx}{(1+\sqrt{x})\left(\sqrt{x-x^2}\right)}$

$=\int \frac{(1-\sqrt{x})dx}{(1-x)\sqrt{x(1-x)}}$

$=\int \frac{1}{(1-x)\sqrt{x(1-x)}}dx-\int \frac{\sqrt{x}dx}{(1-x)\sqrt{x}\sqrt{1-x}}$

Put $(1-x)=t^2$

$\Rightarrow 1-t^2=x$ $dx=-2tdt$

$=\int \frac{-2tdt}{t^2.t.\left(\sqrt{1-t^2}\right)}-\int \frac{-2tdt}{t^2.t}$

$=\int \frac{-2dt}{t^2\sqrt{1-t^2}}+2\int \frac{dt}{t^2}$

$=\int \frac{-2dt}{t^2\sqrt{1-t^2}}-\frac{2}{t}+c$

To evaluate $\int \frac{-2dt}{t^2\sqrt{1-t^2}}$

Put $1-t^2=k^2$

$\Rightarrow -2tdt=2kdk$

$\Rightarrow -2dt=\frac{2kdk}{\sqrt{1-k^2}}$

$\Rightarrow \int \frac{-2dt}{t^2\sqrt{1-t^2}}=\int \frac{2kdk}{\left(\sqrt{1-k^2}\right).(1-k^2).k}$

$=\int \frac{2dk}{{(1-k^2)}^{3/2}}$

$=2.\frac{{(1-k^2)}^{-1/2}}{-1/2}.-2k+C_2$

$=\frac{2k}{\sqrt{1-k^2}}+C_2$

$=\frac{2\sqrt{1-t^2}}{t}+C_2$

$\therefore$ $I=\frac{2\sqrt{1-t^2}}{t}-\frac{2}{t}+C_3$

$=\frac{2\sqrt{x}}{\sqrt{1-x}}-\frac{2}{\sqrt{1-x}}+C_3$

$I=\frac{2\sqrt{x}-2}{\sqrt{1-x}}+C_3=\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+C$