How to prove that $\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}={1\over 4}?$ How do we show that $(1)$
$$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}={1\over 4}?\tag1$$

$$\tan^{-1}(x)=\sum_{n=0}^{\infty}{(-1)^nx^{2n+1}\over 2n+1}$$
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\int_{0}^{\infty}x^{2n+1}\cdot{\mathrm dx\over (1+x)^3}\tag2$$
$$(1+x)^{-3}=\sum_{k=0}^{\infty}{2+k\choose k}x^k$$
$$\sum_{n=0}^{\infty}{(-1)^n\over 2n+1}\sum_{k=0}^{\infty}{2+k\choose k}\int_{0}^{\infty}x^{2n+k+1}\mathrm dx\tag3$$
 A: $$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3}$$
$$\int_{0}^{\infty}\arctan(x)\cdot{\mathrm dx\over (1+x)^3}$$
Apply Integration By Parts $$u = \arctan(x),\,v\prime = \frac{1}{(1 + 3)^3} $$
$$\arctan(x) \left(-\frac{1}{2(1 + 3)^2}\right) -
 \int_{0}^{\infty}\frac{1}{x^2+1}\cdot\left(-\frac{1}{2(1+x)^2}\right)dx$$
$$-\frac{\arctan(x)}{2(1 + x)^2}- \underbrace{\int_{0}^{\infty}-\frac{1}{2(x+1)(x^2 +1)}dx}_I$$
$$I = -\frac{1}{2}\int_{0}^{\infty}-\frac{1}{(x+1)(x^2 +1)}dx$$
Create the partial fraction
$$I = -\frac{1}{2}\int_{0}^{\infty}\left(-\frac{x}{2(x^2 + 1)} + \frac{1}{2(x + 1)}+ \frac{1}{2(x + 1)^2}\right)dx$$
$-\int_{0}^{\infty}\frac{x}{2(x^2 + 1)} dx = -\frac{1}{4}\ln(x^2 + 1)$
$\int_{0}^{\infty}\frac{1}{2(x + 1)}dx= \frac{1}{2}\ln(x + 1)$
$\int_{0}^{\infty}\frac{1}{2(x + 1)^2}dx= -\frac{1}{2(x + 1)}$
$$I = -\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right)$$
Also
$$-\frac{\arctan(x)}{2(1 + x)^2}- \left[ -\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right)\right]$$
$$-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) + C$$

$$\lim_{x \to 0}\left(-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) \right) = -\frac{1}{4}$$
$$\lim_{x \to \infty}\left(-\frac{\arctan(x)}{2(1 + x)^2} +\frac{1}{2}\left(-\frac{1}{4}\ln(x^2 + 1) + \frac{1}{2}\ln(x + 1)-\frac{1}{2(x + 1)}\right) \right) = 0$$


$$\int_{0}^{\infty}\tan^{-1}(x)\cdot{\mathrm dx\over (1+x)^3} = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$$

A: Hint:
Use integration by parts. The derivative of $\arctan x$ is a rational function.
