Question

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

• A. $\frac{1}{2\sqrt{2}}{\tan }^{-1}(\frac{x^2-1}{\sqrt{2}x})-\frac{1}{4\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$
• B. $\frac{1}{2\sqrt{2}}{\tan }^{-1}(\frac{x^2+1}{\sqrt{2}x})+\frac{1}{4\sqrt{2}}\log |\frac{x^2+\sqrt{2}x-1}{x^2-\sqrt{2}x+1}|+c$
• C. $\frac{1}{4\sqrt{2}}{\tan }^{-1}(\frac{x^2+1}{\sqrt{2}x})+\frac{1}{2\sqrt{2}}\log |\frac{x^2+\sqrt{2}x-1}{x^2-\sqrt{2}x+1}|+c$
• D. $\frac{1}{4\sqrt{2}}{\tan }^{-1}(\frac{x^2+1}{\sqrt{2}x})-\frac{1}{2\sqrt{2}}\log |\frac{x^2+\sqrt{2}x-1}{x^2-\sqrt{2}x+1}|+c$

Answer 1

Answer: (A)

$\frac{1}{2\sqrt{2}}{\tan }^{-1}(\frac{x^2-1}{\sqrt{2}x})-\frac{1}{4\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$

$\int \frac{1}{1+x^4}dx=$
$=\frac{1}{2}\int \frac{{1+x}^2-x^2+1}{1+x^4}dx$
$=\frac{1}{2}\int \frac{{1+x}^2}{1+x^4}dx-\frac{1}{2}\int \frac{x^2-1}{1+x^4}dx$
$=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$ (dividing$N^r$and$D^r$by$x^2$)
$=\frac{1}{2}\int \frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+{\left(\sqrt{2}\right)}^2}dx-\frac{1}{2}\int \frac{(1-\frac{1}{x^2})}{{(x+\frac{1}{x})}^2-{\left(\sqrt{2}\right)}^2}dx$
Put $t=x-\frac{1}{x}$ and $u=x+\frac{1}{x}$
$\Rightarrow dt=(1+\frac{1}{x^2})dx$ and $du=(1-\frac{1}{x^2})dx$
$=\frac{1}{2}\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}-\frac{1}{2}\int \frac{du}{u^2-{\left(\sqrt{2}\right)}^2}$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)-\frac{1}{4\sqrt{2}}\log |\frac{u-\sqrt{2}}{u+\sqrt{2}}|+c$
$=\frac{1}{2\sqrt{2}}{\tan }^{-1}(\frac{x^2-1}{\sqrt{2}x})-\frac{1}{4\sqrt{2}}\log |\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}|+c$