Question

$\int \frac{1}{x^4+1}dx$

• A. $=\frac{1}{2\sqrt{2}}{\sec }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{2\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x+1}{x^2+x\sqrt{2}+1}\right|+C$
• B. $=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x+1}{x^2+x\sqrt{2}+1}\right|+C$
• C. $=\frac{1}{4\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x-1}{x^2+x\sqrt{2}-1}\right|+C$
• D. None of these

Answer 1

Answer: (B)

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x+1}{x^2+x\sqrt{2}+1}\right|+C$

$I=\int \frac{1}{x^4+1}dx$
$=\int \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$
$=\frac{1}{2}\int \frac{\frac{2}{x^2}}{x^2+\frac{1}{x^2}}dx$
$=\frac{1}{2}\int (\frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}-\frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}})dx$
$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$
$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+2}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-2}$
Putting $x-\frac{1}{x}=u$ in first integral and $x+\frac{1}{x}=\nu$ in second integral, we get
$I=\frac{1}{2}\int \frac{du}{u^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{2}\int \frac{dv}{{\nu }^2-{\left(\sqrt{2}\right)}^2}$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)-\frac{1}{2}\times \frac{1}{2\sqrt{2}}\log \left|\frac{\nu -\sqrt{2}}{\nu +\sqrt{2}}\right|+C$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{4\sqrt{2}}\log \left|\frac{x^2-\sqrt{2}x+1}{x^2+x\sqrt{2}+1}\right|+C$