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.

$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx +\int_1^{\infty}\frac{\ln x}{(1+x^2)^4}\ dx
\end{align}$
Perform the change of variable $y=\frac{1}{x}$ in the second integral,
$\begin{align}
I&=\int_0^1\ \frac{\ln x}{(1+x^2)^4}dx -\int_1^{\infty}\frac{x^6\ln x}{(1+x^2)^4}\ dx\\
&=\int_0^{1}\frac{(1-x^6)\ln x}{(1+x^2)^4}\ dx\\
&=\left[\frac{x(3x^4+5x^2+3)\ln x}{3(1+x^2)^3}\right]_0^1-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\int_0^1 \frac{3x^4+5x^2+3}{3(1+x^2)^3}\ dx\\
&=-\left[\frac{-x^3+23(1+x^2)^2\arctan x+x}{24(1+x^2)^2}\right]_0^1\\
&=-\frac{23\pi}{96}\\
\end{align}$
