Question

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

• A. $-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+C$
• B. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+C$
• C. $-{\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+C$
• D. None of the above

Answer 1

Answer: (A)

$-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+C$

Let $I=\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$
Put $x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$
Then, $I=\int \frac{-dt}{t^2(1+\frac{1}{t^2})\sqrt{1-{\left(\frac{1}{t}\right)}^2}}$
$=-\int \frac{tdt}{(t^2+1)\sqrt{t^2-1}}$
Again put $t^2-1=z^2\Rightarrow tdt=zdz$
Then, $I=-\int \frac{zdz}{(z^2+2)z}$
$=-\int \frac{1}{z^2+2}dz$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{z}{\sqrt{2}}\right)+C$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{t^2-1}}{\sqrt{2}}\right)+C$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2}x}\right)+C$