How to show $\int_0^\infty \frac{t\ln x}{(x+1)^2+t^2}dx=\frac12\arctan(t)\ln(1+t^2)$

Question

In this post, it mentions the following result without giving a proof.

$$\int_0^\infty \frac{t\ln x}{(x+1)^2+t^2}dx=\frac12\arctan(t)\ln(1+t^2)\tag{1}$$

I know how to compute the case for

$$\int_0^\infty \frac{t\ln x}{x^2+t^2}dx$$

after do the sub $x=ut$,

$$\int_0^\infty \frac{t\ln x}{x^2+t^2}dx=\int_0^\infty \frac{\ln u}{u^2+1}du+\int_0^\infty \frac{\ln t}{u^2+1}du=0+\frac\pi2\ln t$$

But for (1) there is a shift for $x\to x+1$, and we can't use the way we did above. How should I prove this result? Any hint will be appreciated.

Answer 1

Substitute $y=\frac{1+t^2}x$ \begin{align} I=&\int_0^\infty \frac{t\ln x}{(x+1)^2+t^2}dx\\ = & \int_0^\infty \frac{t\ln(1+t^2)-t\ln y}{(y+1)^2+t^2}dy = \int_0^\infty \frac{t\ln(1+t^2)}{(y+1)^2+t^2}dy-I\\ =& \ \frac12\ln(1+t^2) \int_0^\infty \frac{t}{(y+1)^2+t^2}dy = \frac12\ln(1+t^2)\arctan(t) \end{align}