Question

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

• A. $-\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}\right|+c$
• B. $\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}+\sqrt{2}x}\right|+c$
• C. $\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}+c$
• D. $\frac{1}{5\sqrt{2}}\log \left|\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}}\right|+c$

Answer 1

The correct option is A $-\frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}\right|+c$

Putting $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$, we get
$\Rightarrow I=\int \frac{(-\frac{1}{t^2})dt}{(1-\frac{1}{t^2})\sqrt{1+\frac{1}{t^2}}}$

$\Rightarrow I=\int \frac{(-\frac{1}{t^2})\times t^2\times tdt}{(1-t^2)\sqrt{1+t^2}}$

$\Rightarrow I=-\int \frac{tdt}{(t^2-1)\sqrt{t^2+1}}$

Let $t^2+1=u^2$ and $\sqrt{t^2+1}=u$

$\Rightarrow 2tdt=2udu$

$\Rightarrow tdt=udu$

Since, $t^2+1=u^2$

$\Rightarrow t^2+1-2=u^2-2$

$\Rightarrow t^2-1=u^2-2$

Now,

$I=-\int \frac{tdt}{(t^2-1)\sqrt{t^2+1}}$

$\Rightarrow I=-\int \frac{udu}{(u^2-2)u}$

$\Rightarrow I=-\int \frac{du}{(u^2-2)}$
$\therefore I=-\int \frac{du}{u^2-(\sqrt{2}{)}^2}$

$=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{u-\sqrt{2}}{u+\sqrt{2}}\end{array}\right|+c$ [Since, $\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\begin{array}{c}\frac{x-a}{x+a}\end{array}\right|+c$]

$=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{\sqrt{t^2+1}-\sqrt{2}}{\sqrt{t^2+1}+\sqrt{2}}\end{array}\right|+c$

$=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{\sqrt{\frac{1}{x^2}+1}-\sqrt{2}}{\sqrt{\frac{1}{x^2}+1}+\sqrt{2}}\end{array}\right|+c$

$=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{\frac{\sqrt{1+x^2}}{x}-\sqrt{2}}{\frac{\sqrt{1+x^2}}{x}+\sqrt{2}}\end{array}\right|+c$

$=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}\end{array}\right|+c$

$\therefore \int \frac{1}{(1-x^2)\sqrt{1+x^2}}dx=-\frac{1}{2\sqrt{2}}\log \left|\begin{array}{c}\frac{\sqrt{1+x^2}-\sqrt{2}x}{\sqrt{1+x^2}+\sqrt{2}x}\end{array}\right|+c$