Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts , 
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But  it got me no where.
Any hints.
 A: *

*Divide the numerator by $x^2$ and you have $1+\frac{1}{x^2}$

*Divide the denominator by $x^2$ and you have $x^2+\frac{1}{x^2} =\bigl(x-\frac{1}{x}\bigr)^{2}+2$
A: Hint: $$\frac{x^2+1}{x^4+1}=\frac1{2\left(x^2+\sqrt2x+1\right)}+\frac1{2   \left(x^2-\sqrt2x+1\right)}.$$
A: Another approach: Separate out the integral into 
$$
I = \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx
$$
Both take the form of known integral
$$
\int_0^\infty \frac{x^k}{\left(x^n + a\right)^m}\:dx = a^{\frac{k +1}{n} -m}\frac{\Gamma\left(m - \frac{k + 1}{n}\right)\Gamma\left(\frac{k + 1}{n}\right)}{n\Gamma(m)}
$$
if you wish to re-position your integrals into this form, take the following sequences of actions 
(1) Substitute $x \mapsto x^4$
Then
(2) Substitute $x \mapsto \frac{1}{1 + x}$
This will put the integral into the form of the Beta Function 
(3) Use the relationship between the Beta and Gamma Function:
$$
B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}
$$
And you're done. 
For your specific case this then yields
\begin{align*}
I &= \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx \\
&= 1^{\frac{2 +1}{4} -1}\frac{\Gamma\left(1 - \frac{2 + 1}{4}\right)\Gamma\left(\frac{2 + 1}{4}\right)}{4\Gamma(1)}  + 1^{\frac{0 +1}{4} -1}\frac{\Gamma\left(1 - \frac{0 + 1}{4}\right)\Gamma\left(\frac{0 + 1}{4}\right)}{4\Gamma(1)} \\
&= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4}
\end{align*}
Here we employ Euler's Reflection Formula
\begin{equation*}
\Gamma(z)\Gamma(1 - z) = \frac{\pi}{\sin\left(\pi z\right)} \qquad z \not \in \mathbb{Z}
\end{equation*}
Here we have $z = \frac{3}{4}, \frac{1}{4}$ which are both not integers and thus we can use the identity:
\begin{align*}
I &= \frac{\Gamma\left(1 - \frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)}{4} + \frac{\Gamma\left(1 - \frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{4} = \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{3}{4} \right)} + \frac{1}{4} \cdot \frac{\pi}{\sin\left(\pi \cdot \frac{1}{4} \right)} \\
&= \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} + \frac{1}{4} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}} = \frac{\pi}{\sqrt{2}}
\end{align*}
