The easiest way to solve the improper integral $\int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t$? What is the easiest way to solve this improper integral?
$$I = \int_0^{+\infty} \frac{\ln(t)}{(1+t^2)^4} {d}t $$
The value of this integral according to wolfram alpha is $-\frac{23\pi}{96}$. I could find the same result but my method was too long and complicated. I'm looking for a clear and easy method.
 A: Using a semi-circle in the upper half plain avoiding the branch cut and using 
$$f(z) = \frac{\log(z)}{(1+z^2)^4}$$
The contour 

Then we have 
$$\int_{-\infty}^0 \frac{\log|x|+\pi i}{(1+x^2)^4}dx +\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx  = 2\pi i \,\mathrm{Res}(f,i)$$
By change of variable in the first integral 
$$\int_{0}^\infty \frac{\log|x|+\pi i}{(1+x^2)^4}dx +\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx  = 2\pi i \,\mathrm{Res}(f,i)$$
This reduces to 
$$2\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx +\pi i \int_{0}^\infty \frac{1}{(1+x^2)^4}dx  = 2\pi i \,\mathrm{Res}(f,i)$$
Note that we have a pole of order 4 at $z=i$
$$\mathrm{Res}(f,i) = \frac{23 i}{96} + \frac{5 π}{64}$$
By equating the real parts we reach the result 
$$\int_{0}^\infty \frac{\log|x|}{(1+x^2)^4}dx = -\frac{23 \pi}{96}$$
A: Notice that the substitution $t = \tan\theta$ followed by the beta function identity gives
$$ \int_{0}^{\infty} \frac{t^{2a-1}}{(1+t^2)^b} \, dt
= \int_{0}^{\infty} \sin^{2a-1} \theta \cos^{2b-2a-1}\theta \, d\theta
= \frac{1}{2}\frac{\Gamma(a)\Gamma(b-a)}{\Gamma(b)} $$
which is valid for $0 < a < b$. If we plug $b = 4$, then this simplifies to
$$ \int_{0}^{\infty} \frac{t^{2a-1}}{(1+t^2)^4} \, dt
= \frac{(3-a)(2-a)(1-a)\pi}{12\sin(\pi a)} $$
thanks to the Euler's reflection identity. Finally, differentiating both sides w.r.t. $a$ and plugging $a = \frac{1}{2}$ gives the desired result.
A: By considering a keyhole contour, we find that
$$I_1-I_2=2\pi i(\operatorname{Res}(f,i)+\operatorname{Res}(f,-i))=\frac{23\pi^2i}{24}$$
where
\begin{align}f(z)&=\frac{\operatorname{Log}^2(z)}{(1+z^2)^4}
\\I_1&=\int_0^\infty f(x)~\mathrm dx
\\I_2&=\int_0^\infty f(xe^{2\pi i})~\mathrm dx
\end{align}
Where we see that
$$I_1-I_2=\int_0^\infty\frac{\ln^2(z)-(\ln(z)+2\pi i)^2}{(1+z^2)^4}~\mathrm dz=\int_0^\infty\frac{4\pi^2-4\pi i\ln(z)}{(1+z^2)^4}~\mathrm dz$$
And thus,
$$I=-\frac1{4\pi}\Im(I_1-I_2)\implies I=-\frac{23\pi}{96}$$
A: $\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx +\int_1^{\infty}\frac{\ln x}{(1+x^2)^4}\ dx
\end{align}$
Perform the change of variable $y=\frac{1}{x}$ in the second integral,
$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx -\int_1^{\infty}\frac{x^6\ln x}{(1+x^2)^4}\ dx\\
&=\int_0^{1}\frac{(1-x^6)\ln x}{(1+x^2)^4}\ dx\\
&=\left[\frac{x(3x^4+5x^2+3)\ln x}{3(1+x^2)^3}\right]_0^1-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\left[\frac{-x^3+23(1+x^2)^2\arctan x+x}{24(1+x^2)^2}\right]_0^1\\
&=-\frac{23\pi}{96}\\
\end{align}$
