Question

Evaluate

$i)\int \frac{x^2+1}{x^4+1}$ dx
$ii)\int \frac{dx}{x^2+1}$

Answer 1

$i)\int \frac{x^2+1}{x^4+1}dx$

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

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

Let

$x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt$

Therefore, the integral become-

$\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+C$

Substituting the value of $t$, we get

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

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

$ii)\int \frac{1}{x^2+1}dx$

$\int \frac{1}{x^2+{\left(1\right)}^2}dx$

$={\tan }^{-1}x+C$