Question

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

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

Answer 1

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

Given,

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

Let $t=\sqrt{x}$

So, $dt=\frac{1}{2\sqrt{x}}dx$

$\therefore dx=2tdt$

Substituting,

$\int \frac{1}{(t+1)\left(\sqrt{t^2-t^4}\right)}2tdt$

$=2\int \frac{t}{(t+1)\left(\sqrt{t^2-t^4}\right)}dt$

$=2\int \frac{dt}{(t+1)\left(\sqrt{t^2-t^4}\right)}$

substitute $t=\sin \theta$

So, $dt=\cos \theta d\theta$

Solving,

$2\int \frac{\cos \theta d\theta }{(\sin \theta +1)(\sqrt{1-\left({\sin }^2\theta \right)}}$

$=2\int \frac{\cos \theta d\theta }{(\sqrt{{\cos }^2\theta )}}$

as $1-{\sin }^2\theta ={\cos }^2\theta$

$=2\int \frac{\cos \theta d\theta }{(\sin \theta +1)\cos \theta }$

$=2\int \frac{d\theta }{\sin \theta +1}$

$=2\int \frac{1}{\sin \theta +1}\frac{(1-\sin \theta )}{(1-\sin \theta )}d\theta$

$=2\int \frac{1-\sin \theta }{1-{\sin }^2\theta }d\theta$

as $(a+b)(a-b)=a^2-b^2$

$=2\int \frac{1-\sin \theta }{{\cos }^2\theta }d\theta$

$=2\int \frac{1}{{\cos }^2\theta }d\theta -2\int \frac{\sin \theta }{{\cos }^2\theta }d\theta$

$=2\int {\sec }^2\theta d\theta -2\int \tan \theta {\sec }^2\theta d\theta$

$=2\tan \theta -2\sec \theta +c$

$=2\frac{\sin \theta }{\cos \theta }-2\times \frac{1}{\cos \theta }+c$

$=2\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}-2\times \frac{1}{\sqrt{1-{\sin }^2\theta }}+c$

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

$=\frac{2(t-1)}{\sqrt{1-t^2}}+c$

$=\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+c$

So, integration of $\frac{1}{(\sqrt{x-x^2)}(\sqrt{x}+1)}$

is $\frac{2(\sqrt{x}-1)}{\sqrt{1-x}}+c$