How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
 A: Whenever I see an integrand that is sort of symmetric under $x \to x^{-1}$,
I will try to see whether I can express the intergrand in terms of $x \pm x^{-1}$.
Using following identities, 
$$\frac{dx}{x} = \frac{d(x-x^{-1})}{x+x^{-1}} = \frac{d(x+x^{-1})}{x-x^{-1}}$$
we find
$$\begin{align}
\int_0^\infty \frac{x^2+1}{x^4+1}dx 
&= \int_0^\infty \frac{x+x^{-1}}{x^2+x^{-2}} \frac{dx}{x} = \int_0^\infty \frac{x+x^{-1}}{x^2+x^{-2}} \frac{d(x-x^{-1})}{x+x^{-1}} = \int_0^\infty \frac{d(x-x^{-1})}{(x-x^{-1})^2+2}\\
&= \int_{-\infty}^\infty \frac{dy}{y^2+2} = \left[\frac{1}{\sqrt{2}}\tan^{-1}\frac{y}{\sqrt{2}}\right]_{-\infty}^\infty 
= \frac{\pi}{\sqrt{2}}
\end{align}
$$
A: $$\int_{0}^{1}\frac{x^2+1}{x^4+1}\,dx + \int_{1}^{+\infty}\frac{x^2+1}{x^4+1}\,dx = 2\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx = 2\int_{0}^{1}\frac{1+x^2-x^4-x^6}{1-x^8}\,dx$$
hence the value of the integral is provided by the (Dirichlet) series
$$ 2\sum_{k\geq 0}\left(\frac{1}{8k+1}+\frac{1}{8k+3}-\frac{1}{8k+5}-\frac{1}{8k+7}\right) $$
that can be computed through the digamma reflection formula. Its value is just
$$ \boxed{I = \color{red}{\frac{\pi}{\sqrt{2}}}} .$$
The given integral is straightforward to compute through the residue theorem, too.
A: By symmetry,
$$2\int_0^{+\infty}\frac{x^2+1}{x^4+1}\ dx=\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx$$
By taking a semi-circle contour in the upper half of the complex plane and applying Cauchy's residue formula, we find that
$$\begin{align}\int_{-\infty}^{+\infty}\frac{x^2+1}{x^4+1}\ dx&=2\pi i\left[\lim_{a\to e^{\pi i/4}}\frac{(a^2+1)(a-e^{\pi i/4})}{a^4+1}+\lim_{b\to e^{3\pi i/4}}\frac{(b^2+1)(b-e^{3\pi i/4})}{b^4+1}\right]\\&=2\pi i\left[\frac{1+i}{4e^{3\pi i/4}}+\frac{1-i}{4e^{9\pi i/4}}\right]\\&=\frac{\pi i}{2\sqrt2}\left[(1-i)^2-(1+i)^2\right]\\&=\sqrt2\pi\end{align}$$
Thus,

$$\int_0^{+\infty}\frac{x^2+1}{x^4+1}\ dx=\frac{\sqrt2\pi}2=\frac\pi{\sqrt2}$$

A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\Huge\left.a\right)$

With $\ds{p \equiv \exp\pars{{\pi \over 4}\,\ic}}$ and integrating in a quarter circle in the first quadrant:

\begin{align}
\int_{0}^{\infty}{x^{2} + 1 \over x^{4} + 1}\,\dd x & =
\Re\pars{2\pi\ic\,\lim_{x \to p}{\bracks{x\vphantom{^{2}} - p}\bracks{x^{2} + 1} \over x^{4} + 1}}
\label{1}\tag{1}
\\[5mm] & =
\Re\pars{2\pi\ic\,\lim_{x \to p}{x^{3} - px^{2} + x - p \over x^{4} + 1}}
\\[5mm] & =
-2\pi\,\Im\pars{\lim_{x \to p}{3x^{2} - 2px + 1 \over 4x^{3}}} =
-\,{\pi \over 2}\,\Im\pars{\lim_{x \to p}{3x^{3} - 2px^{2} + x \over x^{4}}}
\\[5mm] & =
{\pi \over 2}\,\Im\pars{p^{3} + p} =
{\pi \over 2}\,\Im\pars{p^{2}\bracks{p + {1 \over p}}} =
{\pi \over 2}\,\Im\pars{\ic\bracks{2\cos\pars{\pi \over 4}}}
\\[5mm] & = \bbx{{\root{2} \over 2}\,\pi}
\end{align}

In \eqref{1},
  
  
*
  
*I omitted the integration along the imaginary axis because its real part vanishes out.
  
*The integral along the arc, of radius $\ds{R}$, is of $\ds{\,\mrm{O}\pars{1 \over R}}$ as $\ds{R \to \infty}$ such that it vanishes out in such a limit.


$\Huge\left.b\right)$

With $\ds{\Lambda > 0}$:

\begin{align}
\int_{0}^{\infty}{x^{2} + 1 \over x^{4} + 1}\,\dd x & =
\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{x^{2} + 1 \over x^{4} + 1}\,\dd x = 
\Im\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}{x^{2} + 1 \over x^{2} - \ic}
\,\dd x
\\[5mm] & =
\Im\pars{\bracks{1 + \ic}\lim_{\Lambda \to \infty}
\int_{0}^{\Lambda}{\dd x \over x^{2} - \ic}}
\\[5mm] & =
\Im\pars{{1 + \ic \over 2\root{i}}\lim_{\Lambda \to \infty}\int_{0}^{\Lambda}\bracks{%
{1 \over x - \root{\ic}} - {1 \over x + \root{\ic}}}\dd x}
\\[5mm] & =
{1 \over 2}\,
\Im\pars{\bracks{\expo{-\ic\pi/4} + \expo{\ic\pi/4}}\bracks{\pi\ic}} =
\bbx{{\root{2} \over 2}\,\pi}
\end{align}
A: By substituting $x\mapsto\frac1x$, we get
$$
\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x=\int_1^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
$$
Therefore, using formula $(6)$ from this answer, we get
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
&=2\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x\\
&=2\int_0^1\left(1+x^2-x^4-x^6+x^8+x^{10}-x^{12}-x^{14}+\cdots\right)\,\mathrm{d}x\\
&=2\left(1+\frac13-\frac15-\frac17+\frac19+\frac1{11}-\frac1{13}-\frac1{15}+\cdots\right)\\
&=2\sum_{k=1}^\infty\left(\frac1{8k-7}+\frac1{8k-5}-\frac1{8k-3}-\frac1{8k-1}\right)\\
&=\frac14\sum_{k=1}^\infty\left[\left(\tfrac1k-\tfrac1{k-\frac18}\right)+\left(\tfrac1k-\tfrac1{k-\frac38}\right)-\left(\tfrac1k-\tfrac1{k-\frac58}\right)-\left(\tfrac1k-\tfrac1{k-\frac78}\right)\right]\\
&=\frac14\left(\color{#C00}{H_{-\frac18}}+\color{#090}{H_{-\frac38}-H_{-\frac58}}\color{#C00}{-H_{-\frac78}}\right)\\[4pt]
&=\frac\pi4\left(\color{#C00}{\cot\left(\frac\pi8\right)}+\color{#090}{\cot\left(\frac{3\pi}8\right)}\right)\\[5pt]
&=\frac\pi{\sqrt2}
\end{align}
$$
Since $\cot\left(\frac\pi8\right)=\sqrt2+1$ and $\cot\left(\frac{3\pi}8\right)=\sqrt2-1$.
A: Substituting $x = t^{\frac{1}{4}}$ yields the integral:
$$\frac{1}{4}\int_0^{\infty}\frac{x^{-\frac{1}{4}} + x^{-\frac{3}{4}}}{1+x}dx=\frac{\pi}{2\sin\left(\frac{\pi}{4}\right)}=\frac{\pi}{\sqrt{2}}$$
where we've used the result:
$$\int_0^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin(\pi p)}$$
This can be proven using contour integration, see page 12 of this document.
A: The function integrand is positive.

By equivalence test

Near zero, no problem.
Near $+\infty, $ we have
$$\frac {x^2+1}{x^4+1}\sim \frac {x^2}{x^4}\sim \frac {1}{x^2} .$$
thus it Converges.

Or by comparison test

for great enough $x $
$$\frac {x^2+1}{x^4+1}\leq \frac {2x^2}{x^4} =\frac {2}{x^2}$$
