Question

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

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

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}{\tan }^1\frac{x^2-1}{\sqrt{2}x}+\frac{1}{4\sqrt{2}}\log \left|\frac{x^2+1+\sqrt{2}x}{x^2+1-\sqrt{2}x}\right|+c$

Now,

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

$=\frac{1}{2}\int \frac{(x^2+1)-(x^2-1)}{1+x^4}dx$

$=\frac{1}{2}\int \frac{x^2+1}{1+x^4}dx-$$\frac{1}{2}\int \frac{x^2-1}{1+x^4}dx$

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

$=\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}-$$\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-2}$

$=\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+(\sqrt{2}{)}^2}-$$\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-(\sqrt{2}{)}^2}$

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

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