Question

Evaluate $\int \frac{1}{x^4+5x^2+1}dx$.

• A. $I=\frac{1}{2}[\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{7}}\right)-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{3}}\right)]+C$
• B. $I=\frac{1}{4}[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{7}}\right)-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{3}}\right)]+C$
• C. $I=\frac{1}{2}[\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{7}}\right)+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{3}}\right)]+C$
• D. None of these

Answer 1

The correct option is A $I=\frac{1}{2}[\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{7}}\right)-\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{3}}\right)]+C$

Let $I=\frac{1}{2}\int \frac{2}{x^4+5x^2+1}dx$
$⟹I=\frac{1}{2}\int \frac{1+x^2}{x^4+5x^2+1}dx+\frac{1}{2}\int \frac{1-x^2}{x^4+5x^2+1}dx$
$I=\frac{1}{2}\int \frac{1+\frac{1}{x^2}}{x^2+5+\frac{1}{x^2}}dx-\frac{1}{2}\int \frac{1-\frac{1}{x^2}}{x^2+5+\frac{1}{x^2}}dx$ (dividing $N^r$ and $D^r$ by $x^2$)
$=\frac{1}{2}\int \frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+7}dx-\frac{1}{2}\int \frac{(1-\frac{1}{x^2})}{{(x+\frac{1}{x})}^2+3}dx$
Put $t=x-\frac{1}{x}$ and $u=x+\frac{1}{x}$
$\Rightarrow dt=dx$ and $du=dx$
$=\frac{1}{2}\int \frac{dt}{t^2+{\left(\sqrt{7}\right)}^2}-\frac{1}{2}\int \frac{du}{u^2+{\left(\sqrt{3}\right)}^2}$
$\therefore I=\frac{1}{2}.\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{t}{\sqrt{7}}\right)-\frac{1}{2}.\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{u}{\sqrt{3}}\right)+C$
$=\frac{1}{2}[\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{7}}\right)-\frac{1}{\sqrt{3}}{\tan }^{-1}(\frac{x+\frac{1}{x}}{\sqrt{3}})]+C$