# Find the value of the integral: $\int_{0}^{1}\frac{x}{1+x^4}\arctan(x)\,\mathrm{d}x$

Prove that: $$\int_{0}^{1}\dfrac{x}{1+x^4}\,\arctan(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\,\approx\,0.211322\dots$$

Also prove that: $$\int_{0}^{1}\dfrac{x^3}{1+x^4}\,\text{artanh}(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\,\approx\, 0.211322\dots\,=\,\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x) \,\mathrm{d}x$$

Notice that $$\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)\Longrightarrow$$ $$\frac{1}{2}\,\operatorname{Re}(\arctan^2(\sqrt{\mathrm{i}}))$$

I used a very long and probably wrong method, following the technique used in one of my previous posts: The definite Integral $I=\int_{0}^{1}\frac{x\cdot \operatorname{artanh} x}{1+x^2} dx$

Basically one can recreate this integral by replacing $$x=\sqrt{\mathrm{i}}$$ in the $$\arctan^2(x)$$ expansion and equating the real part to the corresponding infinite G.P generated paralelly... (I’m sorry for being so vague, I’ve been off my medications for a week)

I would greatly appreciate any alternate derivations :)

My derivation:

Consider: $$\arctan(\sqrt{i})=\sqrt{i}-\dfrac{\sqrt{i}^3}{3}+\dfrac{\sqrt{i}^5}{5}-\dfrac{\sqrt{i}^7}{7}+\cdots=\sqrt{i}\left(\left(\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{9}-\cdots\right)-\left(\dfrac{i}{3}-\dfrac{i}{7}+\dfrac{i}{11}-\cdots\right)\right)$$

WKT: $$\arctan(\sqrt{i})=\dfrac{\pi}{4}+i\dfrac{\ln{(\sqrt{2}+1})}{2}$$ and $$\sqrt{i}=\dfrac{i+1}{\sqrt{2}}$$

From these identities it is easy to extrapolate the identities: $$\dfrac{\pi}{4\sqrt{2}}+\dfrac{\ln(\sqrt{2}+1)}{2\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{9}-\dfrac{1}{13}+\cdots$$ And:$$\dfrac{\pi}{4\sqrt{2}}-\dfrac{\ln(\sqrt{2}+1)}{2\sqrt{2}}=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{11}-\dfrac{1}{15}+\cdots$$

Multiplying theses two gives $$\dfrac{\pi^2}{16}-\dfrac{\ln^2{(\sqrt{2}+1)}}{4}=\dfrac{1}{1\cdot3}-\dfrac{1}{1\cdot7}+\cdots-\dfrac{1}{5\cdot3}+\dfrac{1}{5\cdot7}-\cdots$$

This sequence can be alternatively derived from: $$\int_{0}^{1}{\sum_{n=1}^{\infty}(-1)^{n+1}x^{4n+1}\arctan(x) dx}=\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x) dx$$

Therefore,$$\int_{0}^{1}\dfrac{x}{1+x^4}\,\arctan(x)\,\mathrm{d}x\,=\,\left(\dfrac{\pi^2}{16} -\dfrac{\ln^2{(\sqrt{2}+1)}}{4}\right)$$

I might have made a few mistakes somewhere... Also I would like to know if these kinds of integrals have a particular name.

• Numerical integration through integral-calculator suggest $0.211322$ is indeed (almost)correct. Where $\pi^2/16$ gives us $0.616\cdots$ Nov 1, 2023 at 18:49
• Note that $\arctan(\sqrt{i})=(0.42264\cdots)+(0.6922\cdots) i$ Nov 1, 2023 at 18:58
• Ohh sorry :) My bad G. Like I said I'm very absent minded today. I meant to write $\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(x) dx$ Nov 1, 2023 at 19:01
• Very nice derivation. While multiiplying the equations a factor of $\frac12$ was however lost in the LHS.
– user
Dec 1, 2023 at 6:18

Here's a fun proof for the second part of your question. (Your conjecture is off by a factor of $$1/2$$.)

$$\bbox[15px,#E6FFF3 ,border:5px groove #FF8B00 ]{\int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx=\frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)}$$

Proof. Let the integral equal $$\mathcal{I}$$. Then map $$x \mapsto \tanh x$$ so that

\begin{align} \mathcal{I} &:= \int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx \\ &= \int_{0}^{\infty}\frac{x\tanh\left(x\right)^{3}}{1+\tanh\left(x\right)^{4}}\operatorname{sech}^{2}xdx \\ &= \int_{0}^{\infty}\frac{2xe^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx \\ &= \int_{\mathbb{R}}\frac{xe^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx \\ \end{align}

where in the last line, we use the fact that the integrand is an even function.

Let $$z \mapsto \displaystyle \frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)}$$ and define it by the holomorphic function $$\displaystyle f: \mathbb{C} \backslash \left\{n \in \mathbb{Z}: z_1(n), z_2(n), z_3(n)\right\} \to \mathbb{C}$$. The simple poles excluded from the domain are

\begin{align} z_{1}(n)&=\frac{i\pi}{2}\left(2n+1\right) \\ z_{2}(n)&=\frac{1}{4}\ln\left(3-2\sqrt{2}\right)+\frac{i\pi}{4}\left(2n+1\right) \\ z_{3}(n)&=\frac{1}{4}\ln\left(3+2\sqrt{2}\right)+\frac{i\pi}{4}\left(2n+1\right) \end{align}

which are obtained from finding the zeroes of $$\displaystyle \frac{1}{f}$$. We also construct a rectangular box contour $$C$$ with vertices $$(-R,0)$$, $$(R,0)$$, $$(R,R+i\pi)$$, and $$(-R,-R+i\pi)$$ where $$R$$ is sufficiently large such that $$R \gg \displaystyle \frac{1}{4}\ln(3+2\sqrt{2})$$. Below, I provide a visual I made of what $$C$$ looks like in the phase plot of $$f$$ with domain coloring and shading.

We write each contribution of the integral over $$C$$ as

$$\require{cancel}\oint_C f(z)dz=\cancelto{0}{\int_{-R}^{R}f(x)dx}+\int_{R}^{R+i\pi}f(z)dz+\int_{R+i\pi}^{-R+i\pi}f(z)dz+\int_{-R+i\pi}^{-R}f(z)dz$$

where the integral over $$[-R,R]$$ vanishes because the integrand is an odd function.

Taking $$R\to\infty$$ on both sides, we get

$$\require{cancel} \lim_{R\to\infty}\oint_C f(z)dz = \cancelto{0}{\lim_{R\to\infty}\int_{R}^{R+i\pi}f(z)dz} + \lim_{R\to\infty}\int_{R+i\pi}^{-R+i\pi}f(z)dz + \cancelto{0}{\lim_{R\to\infty}\int_{-R+i\pi}^{-R}f(z)dz}\,.$$

The integral over the vertical line segment on the right of $$C$$ vanishes due to the Estimation Lemma and the Squeeze Theorem, and the other integral over the left vertical segment also vanishes for the same kind of reason, I think.

To prove the integral over the vertical line segment on the right of the box decays to $$0$$ as $$R \to \infty$$, we bound the modulus of $$f(z)$$ first like

\begin{align} |f(z)| &= \left|\frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)}\right| \\ &\leq \frac{\left|z\right|^{2}\cdot\left|e^{2z}\right|\cdot\left(3\left|e^{z}\right|+3\left|e^{2z}\right|+3\left|e^{3z}\right|+1\right)^{2}}{\left|\left|e^{2z}\right|-1\right|\cdot\left|6\left|e^{4z}\right|-\left|e^{8z}\right|-1\right|}\,. \\ \end{align}

Parameterizing $$z=R+iy$$ for $$y \in [0,\pi]$$, we get

\begin{align} |f(R+iy)| &\leq \frac{\left|R+iy\right|^{2}\cdot\left|e^{2\left(R+iy\right)}\right|\cdot\left(3\left|e^{\left(R+iy\right)}\right|+3\left|e^{2\left(R+iy\right)}\right|+3\left|e^{3\left(R+iy\right)}\right|+1\right)^{2}}{\left|\left|e^{2\left(R+iy\right)}\right|-1\right|\cdot\left|6\left|e^{4\left(R+iy\right)}\right|-\left|e^{8\left(R+iy\right)}\right|-1\right|} \\ &\leq \frac{e^{2R}\left(R^{2}+2Ry+y^{2}\right)\left(3e^{R}+3e^{2R}+3e^{3R}+1\right)^{2}}{\left(e^{2R}-1\right)\left(e^{8R}-6e^{4R}+1\right)} \\ &:= M(R)\,. \\ \end{align}

As $$R \to \infty$$, we see that $$M(R)\to 0$$ because the largest expression in the numerator grows slower than that of the denominator. Using the Estimation Lemma, we write

$$0 \leq \left|\int_{R}^{R+i\pi} f(z)dz\right| \leq \pi M(R)\,.$$

The $$\pi$$ comes from the length of the right side of the box contour. Taking $$R \to \infty$$ and employing the Squeeze Theorem, we get

$$\lim_{R \to \infty} 0 \leq \lim_{R \to \infty} \left|\int_{R}^{R+i\pi} f(z)dz\right| \leq \pi \lim_{R \to \infty} M(R)$$ $$\implies \lim_{R \to \infty}\left|\int_{R}^{R+i\pi} f(z)dz\right| = 0\,.$$

Thus,

$$\bbox[15px,#FFF5FE]{\lim_{R \to \infty}\int_{R}^{R+i\pi} f(z)dz = 0\,.}$$

The other integral should be evaluated in a similar fashion.

For the integral that doesn't vanish, it surprisingly helps us recover the integral we want. Parameterizing $$z = x+ i\pi$$, we get

\begin{align} \lim_{R\to\infty}\int_{R+i\pi}^{-R+i\pi}f(z)dz &= -\lim_{R\to\infty}\int_{-R}^{R}f(x+i\pi)d(x+i\pi) \\ &= -\lim_{R\to\infty}\int_{-R}^{R}\frac{\left(x+i\pi\right)^{2}e^{2\left(x+i\pi\right)}\left(e^{x+i\pi}-1\right)^{3}\left(e^{x+i\pi}+1\right)^{3}}{\left(e^{2\left(x+i\pi\right)}+1\right)\left(6e^{4\left(x+i\pi\right)}+e^{8\left(x+i\pi\right)}+1\right)}dx \\ \require{cancel} &= -\lim_{R\to\infty}\cancelto{0}{\int_{-R}^{R}f(x)dx} - 2\pi i \mathcal{I} + \pi^2 \lim_{R\to\infty}\cancelto{0}{\int_{-R}^{R}\frac{e^{2x}\left(e^{x}-1\right)^{3}\left(e^{x}+1\right)^{3}}{\left(e^{2x}+1\right)\left(6e^{4x}+e^{8x}+1\right)}dx}\,. \\ \end{align}

The last integral vanishes — thankfully — because its integrand is an odd function. Same for the other one that canceled to $$0$$.

So far, we have

$$-\frac{1}{2\pi i} \lim_{R\to\infty} \oint_C f(z)dz = \mathcal{I}\,.$$

But we can proceed further using Cauchy's Residue Theorem and equating the real part on both sides as follows:

$$-\Re\frac{1}{\cancel{2\pi i}} \cancel{2\pi i} \sum_{n \in \mathcal{P}}\mathop{\mathrm{Res}}_{z=n}f(z) = \Re I = I\,.$$

Here, we will denote $$\mathcal{P}$$ as the set of poles inside $$C$$, which is $$\left\{z_1(0), z_2(0), z_2(1), z_3(0), z_3(1)\right\}$$.

Taking the residue at the simple pole $$z_1(0)$$ of $$f(z)dz$$, we have

\begin{align} -\Re \mathop{\mathrm{Res}}_{z=i\pi/2}f(z) &= -\Re\lim_{z \to i\pi/2}\frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{\frac{d}{dz}\left(e^{2z}+1\right)\left(6e^{4z}+e^{8z}+1\right)} \\ &= -\Re\lim_{z \to i\pi/2} \frac{z^{2}e^{2z}\left(e^{z}-1\right)^{3}\left(e^{z}+1\right)^{3}}{2e^{2z}\left(12e^{2z}+18e^{4z}+4e^{6z}+5e^{8z}+1\right)} \\ &= -\Re \frac{\pi^2}{8} \\ &= -\frac{\pi^2}{8}\,. \\ \end{align}

You can probably guess that the other residues at the four other elements of $$\mathcal{P}$$ are a drag to calculate, so I'll omit the details. Basically,

\begin{align} -\Re \mathop{\mathrm{Res}}_{z=z_2(0)}f(z) &= \frac{\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3-2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_2(1)}f(z) &= \frac{9\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3-2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_3(0)}f(z) &= \frac{\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3+2\sqrt{2}\right) \\ -\Re \mathop{\mathrm{Res}}_{z=z_3(1)}f(z) &= \frac{9\pi^{2}}{128}-\frac{1}{128}\ln^{2}\left(3+2\sqrt{2}\right)\,, \\ \end{align}

after a lot of brutal computations.

All five residues sum to

$$\bbox[15px,border:5px groove #FF18F3 ]{-\Re\sum_{n \in \mathcal{P}}\mathop{\mathrm{Res}}_{z=n}f(z) = \frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)\,.}$$

We finally have

$$\bbox[15px,#FBFFEB ,border:5px groove #00906E ]{\int_{0}^{1}\frac{x^{3}\operatorname{artanh}x}{1+x^{4}}dx=\frac{\pi^{2}}{32}-\frac{1}{8}\ln^{2}\left(1+\sqrt{2}\right)}$$

and we're done! $$\blacksquare$$

Cheers :)

• Absolutely beautiful approach 👍👍. I'll have to learn all the concepts you mentioned here before though :) Nov 6, 2023 at 1:00
• @Borzoi Aw, thank you ❤️❤️ Yes, there are definitely a lot of stuff here that requires some creativity to come up with, but they are worth learning. Nov 6, 2023 at 2:15
• @Accelerator Amazing! Nov 6, 2023 at 17:27
• @ayan Thank you 😊 ❤️ it took a long time to type and make the screenshot Nov 6, 2023 at 17:41
• I don't know if I'm more impressed by your complex analysis or by your equation styling. That's some mathjax skill at work Nov 7, 2023 at 21:55

Integrate by parts, then substitute $$x=\sqrt{\dfrac{1-y}{1+y}}$$ and use the identity $$\arctan\dfrac{1-t}{1+t}=\dfrac\pi4-\arctan t$$ :

\begin{align*} & \int_0^1 \frac x{1+x^4} \arctan x \, dx \\ &= \frac{\pi^2}{32} - \frac12 \int_0^1 \frac{\arctan x^2}{1+x^2} \, dx \\ &= \frac{\pi^2}{32} - \frac14 \int_0^1 \frac{\frac\pi4-\arctan y}{\sqrt{1-y^2}} \, dy \\ &= \frac{\pi^2}{32} - \frac\pi{16} \underbrace{\int_0^1 \frac{dy}{\sqrt{1-y^2}}}_{=\tfrac\pi2} + \frac14 \int_0^1 \frac{\arctan y}{\sqrt{1-y^2}} \, dy \\ &= \frac{\pi^2}{32} + \frac18 \ln\left(\sqrt2+1\right) \ln\left(\sqrt2-1\right) = \boxed{\frac{\pi^2}{32} - \frac18 \operatorname{arcoth}^2\sqrt2} \end{align*}

The last integral is equivalent to $$\int_0^{\frac\pi2}\arctan\left(\sin x\right)\,dx$$.

– user
Nov 30, 2023 at 15:12
• Thanks for catching that @user Nov 30, 2023 at 15:45

We use:

$$\arctan x = \int_0^1 \frac{x dy}{1+x^2 y^2 }$$

Then:

$$I=\int_0^1 \frac{x \arctan x}{1+x^4} dx = \int_0^1 dx \int_0^1 dy \frac{x^2 }{(1+x^4) (1+y^2 x^2)}$$

Introduce:

$$u=x^2,\qquad v=y^2$$

$$I=\frac{1}{4} \int_0^1 du \int_0^1 \frac{dv}{\sqrt{v}} \frac{\sqrt{u}}{(1+u^2) (1+uv)}$$

Partial fractions:

$$\frac{1}{(1+u^2)(1+uv)}=\frac{1-uv}{(1+v^2)(1+u^2)} + \frac{v^2}{(1+v^2)(1+uv)}$$

Then:

$$I= \frac14 (I_1 + I_2 + I_3 )$$

$$I_1 = \int_0^1 du \int_0^1 \frac{dv}{\sqrt{v}} \frac{\sqrt{u}}{(1+u^2) (1+v^2)}$$

$$I_2 = - \int_0^1 du \int_0^1 dv \frac{u^{3/2} \sqrt{v}}{(1+u^2) (1+v^2)}$$

$$I_3 = \int_0^1 du \int_0^1 dv \frac{v^{3/2} \sqrt{u}}{(1+v^2) (1+u v)}= \int_0^1 du \int_0^1 dv \frac{u^{3/2} \sqrt{v}}{(1+u^2) (1+u v)}$$

$$\frac{1}{1+v^2} - \frac{1}{1+uv} = \frac{v(u-v)}{(1+v^2)(1+uv)}$$

$$J = I_2 - I_3 = \int_0^1 \int_0^1 du dv \frac{u^{3/2} v^{3/2} (u-v)}{(1+u^2)(1+v^2) (1+u v)}$$

We have:

$$J=-J \Rightarrow J=0$$

So we obtain:

$$I = \frac{1}{4} I_1$$

$$\int_0^1 du \frac{\sqrt{u}}{1+u^2} = \sum_{n=0}^\infty (-1)^n \int_0^1 u^{2n+1/2} du = \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{n+3/4}=\frac14 \left(\psi \left\{ \frac12 + \frac38 \right\}-\psi \left\{ \frac38 \right\} \right)$$

$$\int_0^1 \frac{dv}{\sqrt{v}} \frac{1}{1+v^2} = \sum_{n=0}^\infty (-1)^n \int_0^1 v^{2n-1/2} du = \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{n+1/4}=\frac14 \left(\psi \left\{ \frac12 + \frac18 \right\}-\psi \left\{ \frac18 \right\} \right)$$

So:

$$\int_0^1 \frac{x \arctan x}{1+x^4} dx = \frac{1}{64} \left(\psi \left\{ \frac58 \right\}-\psi \left\{ \frac18 \right\} \right) \left(\psi \left\{ \frac78 \right\}-\psi \left\{ \frac38 \right\} \right)$$

I don't know how to reduce digamma functions, but the numerical value is correct.

Edit: there's Gauss' digamma theorem (see Wikipedia):

$${\displaystyle \psi \left({\frac {r}{m}}\right)=-\gamma -\ln(2m)-{\frac {\pi }{2}}\cot \left({\frac {r\pi }{m}}\right)+2\sum _{n=1}^{\left\lfloor {\frac {m-1}{2}}\right\rfloor }\cos \left({\frac {2\pi nr}{m}}\right)\ln \sin \left({\frac {\pi n}{m}}\right)}$$

So it's relatively easy to get the answer from here.

• $$\psi\left(\frac58\right)-\psi\left(\frac18\right)=\frac\pi2\left(\sqrt{3-2\sqrt2} + \sqrt{3+2\sqrt2}\right)+2\sqrt2\coth^{-1}\sqrt2$$and$$\psi\left(\frac78\right)-\psi\left(\frac38\right)=\frac\pi2\left(\sqrt{3-2\sqrt2} + \sqrt{3+2\sqrt2}\right)\color{red}{-}2\sqrt2\coth^{-1}\sqrt2$$and their product simplifies nicely enough to$$2\pi^2-8\left(\coth^{-1}\sqrt2\right)^2$$ Nov 1, 2023 at 21:05

Let $$I(a)=\int_{0}^{1}\dfrac{x}{1+x^4}\arctan(ax)\,\mathrm{d}x$$ and then $$\begin{eqnarray} I'(a)&=&\int_{0}^{1}\dfrac{x^2}{(1+a^2x^2)(1+x^4)}\,\mathrm{d}x\\ &=&\frac1{1+a^4}\int_0^1\bigg(\frac{a^2+x^2}{1+x^4}-\frac{a^2}{1+a^2x^2)}\bigg)\,\mathrm{d}x\\ &=&\frac1{1+a^4}\bigg(\int_0^1\frac{a^2+x^2}{1+x^4}\,\mathrm{d}x-a\arctan a\bigg). \end{eqnarray}$$ Now $$\int_0^1\frac{1}{1+x^4}\,\mathrm{d}x=\frac{\pi+2\text{arccoth}(\sqrt 2)}{4\sqrt2}=:A$$ and $$\int_0^1\frac{x^2}{1+x^4}\,\mathrm{d}x=\frac{\pi-2\text{arccoth}(\sqrt 2))}{4\sqrt2}=:B.$$ So $$I'(a)=\frac1{1+a^4}(a^2A+B)-\frac{a\arctan a}{1+a^4}.$$ Therefore $$I(1)=\int_0^1\frac{a^2A+B}{1+a^4}\,\mathrm{d}a-I(1)=2AB-I(1)$$ and hence $$I(1)=AB=\frac{\pi^2}{32}-\frac18\text{arccoth}^2(\sqrt 2).$$

• Wonderful👍👍. Could you try extending this method to $\int_{0}^{1}{x^3 artanh(x)/(1+x^4)}$. I tried it but failed to get the correct answer Nov 30, 2023 at 3:01
• When I have time, I will do. Nov 30, 2023 at 4:08
• I believe you made a small error in the last line. It should be $AB$ instead of $\frac12(A^2+B^2)$ which results also in the change of the sign in the final expression.
– user
Nov 30, 2023 at 11:42
• @user, you’re right. Thanks. Nov 30, 2023 at 13:00

$$\def\arctanh{\operatorname{arctanh}}$$ I very much like the answer given by xpaul, and here the same method will be applied to the calculation of the second integral.

Let $$J(a)=\int_{0}^{1}\dfrac{x^3}{1+x^4}\arctanh(ax)\,\mathrm{d}x.$$ Then $$J(0)=0$$ and $$\begin{eqnarray} J'(a)&=&\int_{0}^{1}\dfrac{x^4}{(1-a^2x^2)(1+x^4)}\,\mathrm{d}x\\ &=&\frac1{1+a^4}\int_0^1\bigg(\frac{1}{1-a^2x^2}-\frac{1+a^2x^2}{1+x^4}\bigg)\,\mathrm{d}x\\ &=&\frac1{1+a^4}\bigg(\frac1a\arctanh a-\int_0^1\frac{1+a^2x^2}{1+x^4}\,\mathrm{d}x\bigg), \end{eqnarray}$$ or $$J'(a)=\bigg(\frac1a-\frac{a^3}{1+a^4}\bigg)\arctanh a-\frac{A+a^2B}{1+a^4},$$ where $$A$$ and $$B$$ have the same meaning as in the above mentioned answer.

So $$J(1)=\int_0^1\frac{\arctanh a}a\,\mathrm{d}a-\int_0^1\frac{A+a^2B}{1+a^4}\,\mathrm{d}a-J(1)$$ and hence $$\begin{eqnarray} J(1)&=&\frac12\bigg(\int_0^1\frac{\arctanh a}a\,\mathrm{d}a-\int_0^1\frac{A+a^2B}{1+a^4}\,\mathrm{d}a\bigg)\\ &=&\frac12\bigg(\frac{\pi^2}8-A^2-B^2\bigg)\\ &=&\boxed{\frac{\pi^2}{32}-\frac{\text{arccoth}^2(\sqrt 2)}8}. \end{eqnarray}$$

• Wonderful answer👍👍. However, the integrals of $xarctanx/(1+x^4)$ and $x^3arctanx/(1+x^4)$ result in the same value from $0->x->1$, 0.2113... So the final answer must match I believe Nov 30, 2023 at 15:59
• @Borzoi They match. xpaul still need to correct the sign in the final expression.
– user
Nov 30, 2023 at 16:02
• Ohhh wonderful 👍👍😊 Nov 30, 2023 at 16:02

$$\int_0^1 \frac{x}{x^4+1}\arctan(x)dx \ \text{ let x=\tan(t)}$$ $$=\int_0^{\frac{\pi}{4}} \frac{t \tan(t) \sec^2(t)}{1+\tan^4(t)} \times \frac{\cos^4(t)}{\cos^4(t)}dt=\int_0^{\frac{\pi}{4}}\frac{t \sin(t) \cos(t)}{\sin^4(t) + \cos^4(t) } dt$$ since $$\sin^2(t) + \cos^2(t)=1$$ then $$\sin^4(t) + \cos^4(t) = 1-2 (\sin(t) \cos(t))^2$$ $$= \int_0^{\frac{\pi}{4}}\frac{t \sin(2t) }{ 2-(\sin(2t)^2} dt \ \ \text{let 2t =u}$$ $$= \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{u \sin(u) }{ 2-\sin(u)^2} du =\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{u \sin(u) }{ 1+\cos(u)^2} du$$ here I will use integration by parts $$\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{u \sin(u) }{ 1+\cos(u)^2} du= \left[ -u \arctan(\cos (u)) \right]_0 ^{\frac{\pi}{2}} + \frac{1}{2}\int_0^{\frac{\pi}{2}} \arctan(\cos(u))du$$ $$=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\arctan(\sin(u))du \color{blue}{\text { by king's rule}}$$ see how to evaluate $$\int_{0}^{\frac{\pi}{2}}\arctan(\sin(x))dx$$