Question

The value of $\int \frac{dx}{(x^2+1)\sqrt{x}}$ equals
$($for some arbitrary constant of integration $C$$)$

• A. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2}x}\right)+\frac{1}{\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$
• B. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2}x}\right)-\frac{1}{\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$
• C. ${\tan }^{-1}\left(\frac{x-1}{\sqrt{2x}}\right)-\frac{1}{\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$
• D. $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2x}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2x}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$

$I=\int \frac{dx}{(x^2+1)\sqrt{x}}$
Put $\sqrt{x}=t\Rightarrow \frac{dx}{\sqrt{x}}=2dt$
Then, $I=\int \frac{2}{t^4+1}dt$
$=\int \frac{(t^2+1)-(t^2-1)}{t^4+1}dt$
$=\int \frac{t^2+1}{t^4+1}dt-\int \frac{t^2-1}{t^4+1}dt$
$=\int \frac{1+1/t^2}{t^2+1/t^2}dt-\int \frac{1-1/t^2}{t^2+1/t^2}dt$

Now, in the first integral, put $z=t-\frac{1}{t}$ and in the second integral, put $u=t+\frac{1}{t}$
Then, $I=\int \frac{dz}{z^2+2}-\int \frac{du}{u^2-2}$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{z}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+C$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t-1/t}{\sqrt{2}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{t+1/t-\sqrt{2}}{t+1/t+\sqrt{2}}\right|+C$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{2x}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{x-\sqrt{2x}+1}{x+\sqrt{2x}+1}\right|+C$