Integrating $\int_{0}^{1} \left(\frac{\arctan(x) - x}{x^2}\right)^2 \,dx$

Question

how to integrate $$\int_{0}^{1} \left(\frac{\arctan(x) - x}{x^2}\right)^2 \,dx$$

Attempt $$=\int_{0}^{1} \left(\frac{\arctan(x) - x}{x^2}\right)^2 \,dx = \int_{0}^{1} \frac{1}{x^4} \cdot (\arctan(x) - x)^2 \,dx$$

Integrating by parts

$$I = -\frac{1}{3} \left(\frac{\pi}{4} - 1\right)^2 + \frac{2}{3} \int_{0}^{1} \frac{x - \arctan(x)}{x(x^2 + 1)} \,dx$$

$$= -\frac{1}{3} \left(\frac{\pi}{4} - 1\right)^2 + \frac{2}{3} \int_{0}^{1} \frac{1}{x^2 + 1} \,dx - \frac{2}{3} \int_{0}^{1} \frac{\arctan(x)}{x(x^2 + 1)} \,dx$$

Answer 1

Apply the differentiation under the integral sign below \begin{align} I(a)=&\int_{0}^{1} \frac{\arctan(a x)}{x(x^2 + 1)}dx\\ I’(a)=& \int_{0}^{1} \frac{1}{(1+x^2)(1+a^2x^2)}dx=\frac{\frac\pi4-a \arctan a}{1-a^2} \end{align} Then, the last integral is \begin{align} &\int_{0}^{1} \frac{\arctan(x)}{x(x^2 + 1)}dx\\ = &\int_0^1 I’(a)da =\int_0^1\frac{\frac\pi4-a \arctan a}{1-a^2}da \\ =&\int_0^1\frac{\frac\pi4- \arctan a}{1-a^2}da + \int_0^1\frac{ \arctan a}{1+a}da\\ \overset{a\to \frac{1-a}{1+a}}=&\ \frac12\int_0^1\frac{ \arctan a}{a}da + \frac12\int_0^1\frac{ \frac\pi4}{1+a}da\\ =& \ \frac12G + \frac\pi8\ln2 \end{align}

Answer 2

In fact, $$\begin{eqnarray} I&=&\int_{0}^{1} \frac{\arctan(x)}{x(x^2 + 1)}dx\\ &=&\int_{0}^{1}\arctan(x)\;d(\ln x-\frac12\ln(1+x^2))\\ &=&\arctan(x)(\ln x-\frac12\ln(1+x^2))\bigg|_0^1-\int_{0}^{1}\frac{\ln x-\frac12\ln(1+x^2)}{1+x^2}\;dx\\ &=&-\frac\pi8\ln2-\int_0^1\frac{\ln x}{1+x^2}\;dx+\frac12\int_0^1\frac{\ln(1+x^2)}{1+x^2}\;dx\\ &=&\frac12G + \frac\pi8\ln2. \end{eqnarray}$$ Here $$\int_0^1\frac{\ln x}{1+x^2}\;dx=-G,\int_0^1\frac{\ln(1+x^2)}{1+x^2}\;dx=-G+\frac\pi2\ln2 $$ are used from https://en.wikipedia.org/wiki/Catalan%27s_constant and Find integral $\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$ (most likely substitution).