$$ I=\int _{ -\infty}^{ \infty} \frac{\log(\lvert t\rvert)}{x^4+t^2} \, dt $$
ATTEMPT:- Since the function is even, the following property may be used:
$$\int_{-a}^a f(x) \, dx=2\int_0^a f(x) \, dx, \text{ if } f(x)=f(-x). \tag 1$$
$$\implies I=2\int_0^\infty \frac{\log(\lvert t\rvert)}{x^4+t^2} \, dt $$
However if we make use of the following property:
$$\int _a^b f(x) \, dx=\int_a^b f(a+b-x) \, dx. \tag 2$$
For the above integral,
$$ I=\int _{-\infty}^\infty \frac{\log(\lvert t\rvert)}{x^4+t^2} \, dt $$
$f(a+b-x)=f(\infty -\infty-x )$ which does not exist.
However on substituting $t=x^2\tan\theta$,
$$ I=\int _{-\frac{\pi}{2}}^{ \frac{\pi}{2}} \frac{\log(\lvert x^2\tan\theta\rvert)}{x^2} \, dt $$
$f(a+b-x)=f(\frac{\pi}{2} -\frac{\pi}{2}-x)$ exists.
Is this just because my integral has changed to other variable?