Question

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

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

Answer 1

Answer: (C)

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

$I=\int \frac{1}{x^4+1}dx=\int (\frac{\sqrt{2}x-2}{4(-x^2+\sqrt{2}x-1)}+\frac{\sqrt{2}x+2}{4(x^2+\sqrt{2}x+1)})dx=\frac{1}{4}\int \frac{\sqrt{2}x+2}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx=\frac{1}{4\sqrt{2}}\int \frac{\sqrt{2}+2x}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{1}{x^2+\sqrt{2}x+1}dx+\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx$
Let $I_1=\frac{1}{4\sqrt{2}}\int \frac{\sqrt{2}+2x}{x^2+\sqrt{2}x+1}dx$
Put $t=x^2+\sqrt{2}x+1\Rightarrow dt=(2x+\sqrt{2})dx$
$I_1=\frac{1}{4\sqrt{2}}\int \frac{1}{t}dt=\frac{\log t}{4\sqrt{2}}=\frac{\log (x^2+\sqrt{2}x+1)}{4\sqrt{2}}$
And $I_2=\frac{1}{4}\int \frac{1}{x^2+\sqrt{2}x+1}dx$
Put $t=x+\frac{1}{\sqrt{2}}\Rightarrow dt=dx$
$I_2=\frac{1}{4}\int \frac{1}{t^2+\frac{1}{2}}dt=\frac{1}{4}\int \frac{2}{2t^2+1}=ta{n}^{-1}\left(\frac{\sqrt{2}x+1}{2\sqrt{2}}\right)I_3=\frac{1}{4}\int \frac{\sqrt{2}x-2}{-x^2+\sqrt{2}x-1}dx=\frac{1}{4}\int (\frac{\sqrt{2}x-2}{\sqrt{2}(-x^2+\sqrt{2}x-1)}-\frac{1}{-x^2+\sqrt{2}x-1})dx=-\frac{log(-x^2+\sqrt{2}x-1)}{4\sqrt{2}}-\frac{tan(1-\sqrt{2}x)}{2\sqrt{2}}{I=I}_1+I_2+I_3=\frac{1}{2\sqrt{2}}[ta{n}^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)-\frac{1}{2}ln\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}]$