∫1+x2dxx4+1 [12√2tan−1x−1x√2]+C [1√2tan−1x−1x2√2]+C [1√

SOLUTION

Let I = $\int \frac{1+x^{2}dx}{x^4+1}$
Let's divide $x^2$ in the numerator or denominator - 
$I=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$
Or $I=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x}{)}^2+2}dx$
Let’s  call this integral as $I$
So, $I=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x}{)}^2+2}dx$
Substitute $x-\frac{1}{x}$ = t
& $(1+\frac{1}{x^2})$ dx = dt
$I=\int \frac{1}{(t{)}^2+({\sqrt{2}}^2)}$ dt
Using the standard formulae we can say
$I=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{t}{\sqrt{2}}\right)+C_1$
Or $I=\frac{1}{\sqrt{2}}ta{n}^{-1}\left(\frac{\frac{x-1}{x}}{\sqrt{2}}\right)+C_1$
So, the final answer will be - 
$=\left[\frac{1}{\sqrt{2}}ta{n}^{-1}\right(\frac{\frac{x-1}{x}}{\sqrt{2}}\left)\right]+C$