Prove $\int_{0}^{\infty} \frac{\arctan(x^3)}{x^2} \ dx = \frac{\pi}{\sqrt3}$

Question

I came up with this integral:

$$ \int_{0}^{\infty} \frac { \arctan(x^3) } { x^2 } \ dx $$

but I couldn't find any way to solve it.

Desmos gave a convergent value of $1.813799...$ which I verified with WolframAlpha to be the exact value $\frac{\pi}{\sqrt{3}}$. Unfortunately Wolfram seems to no longer offer any meaningful preview of step-by-step solutions, so I couldn't glean any help from it for where I could start.

I've only been trying to solve the indefinite integral so far, since I can't spot any tricks for evaluating the definite integral (probably down to lack of experience).

I've tried straight-up parts:

$$ \begin{align*} &\ \int \frac{\arctan\left(x^{3}\right)}{x^{2}} \ dx \\ =&\ \arctan\left(x^{3}\right)\left(-\frac{1}{x}\right) - \int \frac{3x^{2}}{x^{6}+1}\left(-\frac{1}{x}\right) \ dx \\ =&\ -\frac{\arctan\left(x^{3}\right)}{x} + 3\int \frac{x}{x^{6}+1} \ dx \end{align*} $$

but despite obtaining a rational function I don't actually see any way to solve this one either.

Then I tried substitution:

$$ \begin{align*} \arctan\left(x^{3}\right) &= t \\ x^{3} &= \tan t \\ x &= \left(\tan t\right)^{1/3} \\ dx &= \frac{1}{3}\left(\tan t\right)^{-2/3}\cdot\sec^{2}t \ dt \end{align*} $$

which gives

$$ \begin{align*} &\ \int \frac{\arctan\left(x^{3}\right)}{x^{2}} \ dx \\ =&\ \int \frac{t}{\left(\tan t\right)^{2/3}}\cdot\frac{1}{3}\left(\tan t\right)^{-2/3}\cdot\sec^{2}t\ dt \\ =&\ \frac{1}{3} \int \frac{t}{\left(\tan t\right)^{4/3}}\cdot\sec^{2}t\ dt \end{align*} $$

which doesn't look that promising, although trying parts again ($1/3$ omitted for brevity)

$$ \begin{align*} &= t\cdot\bigg(-\frac{3}{\left(\tan t\right)^{1/3}}\bigg)-\int -\frac{3}{\left(\tan t\right)^{1/3}}\ dt \\ &= -\frac{3t}{\left(\tan t\right)^{1/3}}+3 \int \frac{1}{\left(\tan t\right)^{1/3}} \end{align*} $$

I barely know how to solve $\sqrt{\tan{x}}$, so I have no clue how to tackle a fractional exponent.

I would be grateful if you could provide some hints/tips/insights before revealing the method (if there is one), so that I can still try to discover the solution myself before reading your full answer.

Thanks for your time. First post here, so please let me know if there are any issues and how I can improve!

Answer 1

You've made an excellent start with integration by parts!

Let the integral be $\color{navy}{\mathcal{I}}$. We want to evaluate $$\bbox[15px, #E0F8FF, border:5px groove #007BFF]{\color{navy}{\mathcal{I}} := \int_{0}^{\infty} \frac{\arctan(x^3)}{x^2} \ dx}$$

Using integration by parts with $u_0 = \arctan(x^3)$ and $dv_0 = \frac{1}{x^2} dx$. Then $du_0 = \frac{3x^2}{1+x^6} dx$ and $v_0 = -\frac{1}{x}$. So, $$\begin{align*} \color{navy}{\mathcal{I}} &= \left[ -\frac{\arctan(x^3)}{x} \right]_0^\infty - \int_0^\infty \left(-\frac{1}{x}\right) \frac{3x^2}{1+x^6} \ dx \\ &= \left[ -\frac{\arctan(x^3)}{x} \right]_0^\infty + 3 \int_0^\infty \frac{x}{1+x^6} \ dx \end{align*}$$

Let's evaluate the boundary term:

• As $x \to \infty$: $\arctan(x^3) \to \frac{\pi}{2}$. So, $-\frac{\arctan(x^3)}{x} \to -\frac{\pi/2}{x} \to 0$.
• As $x \to 0^+$: We can use the Taylor series for $\arctan(y) \approx y$ for small $y$. So, $\arctan(x^3) \approx x^3$. Thus, $-\frac{\arctan(x^3)}{x} \approx -\frac{x^3}{x} = -x^2 \to 0$.

Therefore, the boundary term is $\require{cancel}\cancelto{0}{\left[ -\frac{\arctan(x^3)}{x} \right]_0^\infty} = 0 - 0 = 0$. The integral simplifies to: $$\color{navy}{\mathcal{I}} = 3 \int_0^\infty \frac{x}{1+x^6} \ dx$$

Let $\color{teal}{\mathcal{J}_1} = \int_0^\infty \frac{x}{1+x^6} \ dx$. So, $\color{navy}{\mathcal{I}} = 3 \color{teal}{\mathcal{J}_1}$.

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Hints for solving $\displaystyle \color{teal}{\mathcal{J}_1} = \int_0^\infty \frac{x}{1+x^6} \ dx$:

1. Substitution: The term $x^6$ suggests a substitution. Try $u = x^2$ (or $x \mapsto \sqrt{u}$). How does the integral transform? You should arrive at an integral of the form $C \int_0^\infty \frac{1}{1+u^3} du$. Let's call this $J_2 = \int_0^\infty \frac{1}{1+u^3} du$.

2. Symmetry/Reciprocal Substitution for $J_2$: For integrals of the form $\int_0^\infty f(u) du$, the substitution $u \mapsto 1/t$ (so $du = -1/t^2 dt$) is often very powerful. Apply this to $J_2$. You should find that $J_2$ is also equal to $\int_0^\infty \frac{t}{1+t^3} dt$.

3. Combine and Simplify: Using the result from hint 2, you have two expressions for $J_2$. Consider $2J_2 = \int_0^\infty \frac{1}{u^3+1} du + \int_0^\infty \frac{u}{u^3+1} du = \int_0^\infty \frac{1+u}{u^3+1} du$. Simplify the integrand $\frac{1+u}{u^3+1}$ by factoring the denominator.

4. Evaluate the Simplified Integral: The simplified integral should be of the form $\int_0^\infty \frac{1}{au^2+bu+c} du$. Complete the square in the denominator. This will lead to an $\arctan$ function.

5. Back-substitute: Once you find $J_2$, remember that $I = 3 J_1$ and $J_1 = \frac{1}{2} J_2$.

Try working through these hints. The full solution follows below.

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Full Solution:

Following the hints:

1. We have $\color{navy}{\mathcal{I}} = 3 \color{teal}{\mathcal{J}_1} = 3 \int_0^\infty \frac{x}{1+x^6} \ dx$. Let $u = x^2$. Then $du = 2x \ dx$, so $x \ dx = \frac{1}{2} du$. The limits remain $0$ to $\infty$. $$\color{teal}{\mathcal{J}_1} = \int_0^\infty \frac{1}{1+u^3} \cdot \frac{1}{2} du = \frac{1}{2} \int_0^\infty \frac{1}{1+u^3} \ du$$ Let $$\displaystyle \bbox[10px, #F0FFF0, border:3px dashed #28A745]{\color{teal}{\mathcal{J}_2} := \int_0^\infty \frac{1}{1+u^3} \ du}$$ so that $\color{navy}{\mathcal{I}} = 3 \color{teal}{\mathcal{J}_1} = \frac{3}{2} \color{teal}{\mathcal{J}_2}$.

2. To evaluate $\color{teal}{\mathcal{J}_2}$, let $u \mapsto 1/t$, so $du = -1/t^2 dt$. When $u=0$, $t=\infty$. When $u=\infty$, $t=0$. $$\color{teal}{\mathcal{J}_2} = \int_\infty^0 \frac{1}{1+(1/t)^3} \left(-\frac{1}{t^2}\right) dt = \int_0^\infty \frac{1}{\frac{t^3+1}{t^3}} \frac{1}{t^2} dt = \int_0^\infty \frac{t^3}{t^3+1} \frac{1}{t^2} dt = \int_0^\infty \frac{t}{t^3+1} dt$$ Since this is a definite integral, the variable name doesn't matter, so $\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{u}{u^3+1} du$.

3. Now we have two expressions for $\color{teal}{\mathcal{J}_2}$: $$\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{u^3+1} du \quad \text{and} \quad \color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{u}{u^3+1} du$$ Adding them gives: $$2\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{u^3+1} du + \int_0^\infty \frac{u}{u^3+1} du = \int_0^\infty \frac{1+u}{u^3+1} du$$ We know $u^3+1 = (u+1)(u^2-u+1)$. So, for $u \ge 0$ (and $u \neq -1$, which is true on our domain of integration): $$\frac{1+u}{u^3+1} = \frac{\cancel{(1+u)}}{\cancel{(u+1)}(u^2-u+1)} = \frac{1}{u^2-u+1}$$

4. Thus, $$2\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{u^2-u+1} du$$ Complete the square in the denominator: $u^2-u+1 = \left(u - \frac{1}{2}\right)^2 - \frac{1}{4} + 1 = \left(u - \frac{1}{2}\right)^2 + \frac{3}{4}$. $$2\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{\left(u - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} du$$ This is a standard arctangent integral: $\int \frac{1}{y^2+a^2} dy = \frac{1}{a}\arctan\left(\frac{y}{a}\right)$. Here $y = u-1/2$ and $a = \sqrt{3}/2$. $$\begin{align*} 2\color{teal}{\mathcal{J}_2} &= \left[ \frac{1}{\sqrt{3}/2} \arctan\left(\frac{u-1/2}{\sqrt{3}/2}\right) \right]_0^\infty \\ &= \left[ \frac{2}{\sqrt{3}} \arctan\left(\frac{2u-1}{\sqrt{3}}\right) \right]_0^\infty \\ &= \frac{2}{\sqrt{3}} \left( \lim_{u\to\infty} \arctan\left(\frac{2u-1}{\sqrt{3}}\right) - \arctan\left(\frac{-1}{\sqrt{3}}\right) \right) \\ &= \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} - \left(-\frac{\pi}{6}\right) \right) \\ &= \frac{2}{\sqrt{3}} \left( \frac{\pi}{2} + \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} \left( \frac{3\pi+\pi}{6} \right) = \frac{2}{\sqrt{3}} \left(\frac{4\pi}{6}\right) = \frac{2}{\sqrt{3}} \left(\frac{2\pi}{3}\right) = \color{maroon}{\frac{4\pi}{3\sqrt{3}}} \end{align*}$$

5. So, $\color{teal}{\mathcal{J}_2} = \frac{1}{2} \cdot \color{maroon}{\frac{4\pi}{3\sqrt{3}}} = \color{maroon}{\frac{2\pi}{3\sqrt{3}}}$. We can state this intermediate result beautifully: $$\bbox[15px, #E6FFF3, border:5px groove #1E90FF]{\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{1+u^3} \ du = \color{maroon}{\frac{2\pi}{3\sqrt{3}}}}$$ Finally, substitute back into $\color{navy}{\mathcal{I}} = \frac{3}{2} \color{teal}{\mathcal{J}_2}$: $$\color{navy}{\mathcal{I}} = \frac{3}{2} \cdot \color{maroon}{\frac{2\pi}{3\sqrt{3}}} = \color{maroon}{\frac{\pi}{\sqrt{3}}}$$ Rationalizing the denominator gives $\color{maroon}{\frac{\pi\sqrt{3}}{3}}$.

Thus, $$\bbox[20px, #FFF0F5, border:5px groove #DB7093 ]{\int_{0}^{\infty} \frac{\arctan(x^3)}{x^2} \ dx = \color{maroon}{\frac{\pi}{\sqrt{3}}}}$$

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Alternative for $\displaystyle \color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{1+u^3} du$ using Partial Fractions:

One could also evaluate $\color{teal}{\mathcal{J}_2}$ using partial fractions. We set up the decomposition: $$\frac{1}{u^3+1} = \frac{1}{(u+1)(u^2-u+1)} = \frac{\color{purple}{A}}{u+1} + \frac{\color{purple}{B}u+\color{purple}{C}}{u^2-u+1}$$

Multiplying by $(u+1)(u^2-u+1)$ gives $1 = \color{purple}{A}(u^2-u+1) + (\color{purple}{B}u+\color{purple}{C})(u+1)$.

To find $\color{purple}{A}, \color{purple}{B}, \color{purple}{C}$:

• For $u=-1$: $1 = \color{purple}{A}(1-(-1)+1) + (\color{purple}{B}(-1)+\color{purple}{C})(-1+1) = \color{purple}{A}(3) + 0 \implies 3\color{purple}{A} = 1 \implies \boldsymbol{\color{purple}{A}=1/3}$.
• Comparing coefficients of $u^2$: $0 \cdot u^2 = \color{purple}{A} u^2 + \color{purple}{B} u^2 \implies \color{purple}{A}+\color{purple}{B} = 0 \implies \color{purple}{B} = -\color{purple}{A} \implies \boldsymbol{\color{purple}{B} = -1/3}$.
• Comparing constant terms: $1 = \color{purple}{A}(1) + \color{purple}{C}(1) \implies \color{purple}{A}+\color{purple}{C} = 1 \implies \color{purple}{C} = 1-\color{purple}{A} = 1-1/3 \implies \boldsymbol{\color{purple}{C} = 2/3}$.

So, $\displaystyle \frac{1}{u^3+1} = \frac{1}{3(u+1)} + \frac{-\frac{1}{3}u+\frac{2}{3}}{u^2-u+1} = \frac{1}{3}\left(\frac{1}{u+1} - \frac{u-2}{u^2-u+1}\right)$.

The indefinite integral is: $$\begin{align*} \int \frac{1}{1+u^3}du &= \frac{1}{3}\int \left(\frac{1}{u+1} - \frac{u-1/2-3/2}{(u-1/2)^2+3/4}\right)du \\ &= \frac{1}{3}\left(\ln|u+1| - \int \frac{u-1/2}{(u-1/2)^2+3/4}du + \frac{3}{2}\int \frac{1}{(u-1/2)^2+(\sqrt{3}/2)^2}du\right) \\ &= \frac{1}{3}\left(\ln|u+1| - \frac{1}{2}\ln(u^2-u+1) + \frac{3}{2} \frac{1}{\sqrt{3}/2}\arctan\left(\frac{u-1/2}{\sqrt{3}/2}\right)\right)+K \\ &= \frac{1}{3}\left(\ln|u+1| - \frac{1}{2}\ln(u^2-u+1) + \sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right)\right)+K \\ &= \frac{1}{3} \left[ \ln\left(\frac{|u+1|}{\sqrt{u^2-u+1}}\right) + \sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right) \right]+K \end{align*}$$

Evaluating the definite integral $\color{teal}{\mathcal{J}_2} = \int_0^\infty \frac{1}{1+u^3}du$: $$\color{teal}{\mathcal{J}_2} = \frac{1}{3} \left[ \ln\left(\frac{u+1}{\sqrt{u^2-u+1}}\right) + \sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right) \right]_0^\infty$$

For the logarithmic term: As $u \to \infty$, $\ln\left(\frac{u+1}{\sqrt{u^2-u+1}}\right) = \ln\left(\frac{u(1+1/u)}{\sqrt{u^2(1-1/u+1/u^2)}}\right) = \ln\left(\frac{\cancel{u}(1+1/u)}{\cancel{u}\sqrt{1-1/u+1/u^2}}\right) \to \ln(1) = 0$.

At $u=0$, $\ln\left(\frac{1}{\sqrt{1}}\right) = \ln(1) = 0$.

So the logarithmic term contributes $\cancelto{0}{0-0}=0$ to the definite integral.

For the arctangent term: $\sqrt{3}\left(\lim_{u\to\infty}\arctan\left(\frac{2u-1}{\sqrt{3}}\right) - \arctan\left(\frac{-1}{\sqrt{3}}\right)\right) = \sqrt{3} \left(\frac{\pi}{2} - \left(-\frac{\pi}{6}\right)\right) = \sqrt{3}\left(\frac{2\pi}{3}\right)$.

So, $$\color{teal}{\mathcal{J}_2} = \frac{1}{3} \left(0 + \sqrt{3} \frac{2\pi}{3}\right) = \color{maroon}{\frac{2\pi\sqrt{3}}{9}} = \color{maroon}{\frac{2\pi}{3\sqrt{3}}}$$

This matches the result obtained by the symmetry trick.

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Regarding your substitution $\boldsymbol{x^3 = \tan t}$ (Beta Function Approach):

This is also an elegant and valid approach. If $x^3 = \tan t$, then $x = (\tan t)^{1/3}$. This implies $dx = \frac{1}{3}(\tan t)^{(1/3)-1} \sec^2 t \ dt = \frac{1}{3}(\tan t)^{-2/3}\sec^2 t \ dt$.

The limits of integration change: as $x \to 0$, $\tan t \to 0 \implies t \to 0$. As $x \to \infty$, $\tan t \to \infty \implies t \to \pi/2$.

The integral becomes: $$\color{navy}{\mathcal{I}} = \int_0^{\pi/2} \frac{t}{((\tan t)^{1/3})^2} \cdot \frac{1}{3}(\tan t)^{-2/3}\sec^2 t \ dt = \frac{1}{3} \int_0^{\pi/2} t (\tan t)^{-4/3} \sec^2 t \ dt$$

We integrate by parts. Let $u_{part}=t$ and $dv_{part}=(\tan t)^{-4/3}\sec^2 t \ dt$. Then $du_{part}=dt$. To find $v_{part}$, let $w = \tan t$, so $dw = \sec^2 t \ dt$. $v_{part} = \int w^{-4/3} dw = \frac{w^{-4/3+1}}{-4/3+1} = \frac{w^{-1/3}}{-1/3} = -3(\tan t)^{-1/3}$.

$$\begin{align*} \color{navy}{\mathcal{I}} &= \frac{1}{3} \left( \left[ t \cdot \left(-3(\tan t)^{-1/3}\right) \right]_0^{\pi/2} - \int_0^{\pi/2} \left(-3(\tan t)^{-1/3}\right) dt \right) \\ &= \cancelto{0}{\frac{1}{3} \left[ -3t(\tan t)^{-1/3} \right]_0^{\pi/2}} + \int_0^{\pi/2} (\tan t)^{-1/3} dt \end{align*}$$

The boundary term vanishes:

• As $t \to (\pi/2)^-$, $t \to \pi/2$ and $\tan t \to \infty$, so $(\tan t)^{-1/3} \to 0$. The expression $-t(\tan t)^{-1/3} \to 0$.
• As $t \to 0^+$, $-t(\tan t)^{-1/3} = -t\left(\frac{\sin t}{\cos t}\right)^{-1/3} = -t (\cos t)^{1/3} (\sin t)^{-1/3}$. Since $\cos t \to 1$ and $\sin t \approx t$ for small $t$, this is $\approx -t \cdot 1 \cdot t^{-1/3} = -t^{2/3} \to 0$.

Thus, $\color{navy}{\mathcal{I}} = \int_0^{\pi/2} (\tan t)^{-1/3} dt = \int_0^{\pi/2} \cot^{1/3}(t) dt$.

This integral can be evaluated using the Beta function. Recall that $\int_0^{\pi/2} (\sin t)^a (\cos t)^b dt = \frac{1}{2} B\left(\frac{a+1}{2}, \frac{b+1}{2}\right)$.

We have $\cot^{1/3}(t) = (\cos t)^{1/3} (\sin t)^{-1/3}$. So $a=-1/3$ and $b=1/3$.

$$\begin{align*} \color{navy}{\mathcal{I}} &= \int_0^{\pi/2} (\sin t)^{-1/3} (\cos t)^{1/3} dt \\ &= \frac{1}{2} B\left(\frac{-1/3+1}{2}, \frac{1/3+1}{2}\right) \\ &= \frac{1}{2} B\left(\frac{2/3}{2}, \frac{4/3}{2}\right) = \frac{1}{2} B\left(\frac{1}{3}, \frac{2}{3}\right) \end{align*}$$

Using the relationship $B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ and Euler's reflection formula $\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$:

$$\color{navy}{\mathcal{I}} = \frac{1}{2} \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1/3+2/3)} = \frac{1}{2} \frac{\Gamma(1/3)\Gamma(1-1/3)}{\Gamma(1)}$$

Since $\Gamma(1)=1$ and taking $z=1/3$ in the reflection formula: $$\color{navy}{\mathcal{I}} = \frac{1}{2} \cdot \frac{\pi}{\sin(\pi/3)} = \frac{1}{2} \cdot \frac{\pi}{\sqrt{3}/2} = \color{maroon}{\frac{\pi}{\sqrt{3}}}$$

This confirms the result, arriving at the same beautiful conclusion: $$\bbox[15px, #F5EEFF, border:5px groove #8A2BE2 ]{\color{navy}{\mathcal{I}} = \color{maroon}{\frac{\pi}{\sqrt{3}}}}$$

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And we're done! $\blacksquare$

Cheers :)

Answer 2

The integral you obtain via parts:

$$\int_0^\infty \frac{x}{1+x^6} \, dx$$

is in standard beta function form:

$$\int_0^\infty \frac{x^{m-1}}{1+x^n} \, dx = \frac{\pi}{n \sin(\frac{m \pi}{n})},$$

which is valid for $0<m<n$. Can you take it from here?

A more direct approach from there is to let $u=x^2$ and use partial fraction decomposition:

$$\frac{1}{1+u^3} = \frac{1}{3(u+1)} + \frac{-u+2}{3(u^2-u+1)}.$$

This works out nicely - you want to complete the square on the bottom of the rightmost term there so you can make use of the general $\arctan$ antiderivative.

Answer 3

Via integration by parts, we have $$ \begin{aligned} \int_0^{\infty} \frac{\arctan \left(x^3\right)}{x^2} d x = & -\int_0^{\infty} \arctan \left(x^3\right) d\left(\frac{1}{x}\right) \\ = & \int_0^{\infty} \frac{1}{x} \cdot \frac{3 x^2}{1+x^6} d x \\ = & 3 \int_0^{\infty} \frac{x}{1+x^6} d x \\ = & \frac{3}{2} \int_0^{\infty} \frac{d x}{1+x^3}, \quad \textrm{ via } x^2 \rightarrow x \end{aligned} $$ Splitting integration interval, we get

$$ \begin{aligned} I & =\frac{3}{2}\left[\int_0^1 \frac{d x}{1+x^3} +\int_1^{\infty} \frac{d x}{1+x^3}\right]\\ & =\frac{3}{2}\int_0^1 \frac{1+x}{1+x^3} d x, \quad \textrm{ via } x\to \frac{1}{x} \\ & =\frac{3}{2} \int_0^1 \frac{1}{1-x+x^2} d x \\ & =6 \int_0^1 \frac{1}{(2 x-1)^2+3} d x \\ & =\sqrt{3}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^1\\&=\frac{\pi}{\sqrt{3}} \end{aligned} $$

Answer 4

Consider $x \mapsto 1/x$ and the integral becomes $$I = \int_\infty^0{\frac{\arctan(1/x^3)}{1/x^2}\cdot(-1/x^2)}dx = \int_0^\infty{\arctan(1/x^3)}dx$$

Now, do IBP with $u = \arctan(1/x^3)$ and $dv = 1$; $$I = \int_0^\infty{\arctan(1/x^3)}dx =\big[x\arctan(1/x^3)\big]_0^\infty + \int_0^\infty{\frac{3x^3}{x^6 + 1}}dx = \int_0^\infty{\frac{3x^3}{x^6 + 1}}dx \tag{1}\label{1}$$

While it is possible to use $u$-subtitution on $\eqref{1}$, the outcome will be tedious. In the motivation to have little terms as possible and obtain a simpler method, we first observe the last term in your IBP:

$$ \begin{align*} &\ I = \int \frac{\arctan\left(x^{3}\right)}{x^{2}} \ dx \\ =&\ \arctan\left(x^{3}\right)\left(-\frac{1}{x}\right) - \int \frac{3x^{2}}{x^{6}+1}\left(-\frac{1}{x}\right) \ dx \\ =&\ -\frac{\arctan\left(x^{3}\right)}{x} + \color\green{3\int \frac{x}{x^{6}+1} \ dx} \end{align*} $$

Particularly, we have $$I = \left[-\frac{\arctan\left(x^{3}\right)}{x}\right]_0^\infty + \int_0^\infty \frac{3x}{x^{6}+1} \ dx = \int_0^\infty \frac{3x}{x^{6}+1} \ dx$$

$$\therefore I + I = \eqref{1} + \int_0^\infty \frac{3x}{x^{6}+1} \ dx \implies I = \frac{1}{2}\int_0^\infty \frac{3x^3 + 3x}{x^{6}+1}dx = \frac{1}{2}\int_0^\infty \frac{3x(x^2 + 1)}{x^{6}+1}dx$$

By cubic factoring we know $(a^3 + 1) = (a + 1)(a^2 - a + 1)$. Let $a = x^2$ or $a^3 = x^6$ to factor $(x^6 + 1) = (x^2 + 1)(x^4 - x^2 + 1)$. Substitute back: $$\frac{1}{2}\int_0^\infty \frac{3x(x^2 + 1)}{(x^2 + 1)(x^4 - x^2 + 1)}dx = \frac{1}{2}\int_0^\infty \frac{3x}{x^4 - x^2 + 1}dx$$

Complete the square on the denominator term w.r.t to $x^2$. That is, $(x^2)^2 - x^2 + 1 = \left(x^2 - 1/2\right)^2 + 3/4$

$$\frac{1}{2}\int_0^\infty \frac{3x}{x^4 - x^2 + 1}dx = \frac{1}{2}\int_0^\infty \frac{3x}{(x^2 - 1/2)^2 + 3/4}dx$$

This is a convenient $u-$sub with $u = x^2 - 1/2,\, du = 2x\,dx$ and leads to a classic integral,

$$\frac{3}{4}\int_{-1/2}^\infty \frac{du}{u^2 + \sqrt{3/4}^2} = \frac{3}{4}\left[\frac{2}{\sqrt{3}}\arctan{\frac{u}{\sqrt{3}/2}}\right]_{-1/2}^\infty = \frac{\sqrt{3}\pi}{4} - \frac{\sqrt{3}}{2}\arctan{\frac{-1/2}{\sqrt{3}/2}} = \frac{\sqrt{3}\pi}{4} + \frac{\sqrt{3}\pi}{12}$$

The result is the same but in different form: $\frac{\sqrt{3}\pi}{4} + \frac{\sqrt{3}\pi}{12} = \frac{3\sqrt{3}\pi}{12} + \frac{\sqrt{3}\pi}{12} = \frac{\sqrt{3}\pi}{3} = \pi/\sqrt{3}$