We are asked to evaluate the following integral: $\int_0^{+\infty} \frac{dx}{(1+x)(\pi^2 + \ln^2 x)}$

1. Problem Description

We are asked to evaluate the following integral:

$\int_0^{+\infty} \frac{dx}{(1+x)(\pi^2 + \ln^2 x)}$

2. Solution Steps

Let $I = \int_0^{\infty} \frac{dx}{(1+x)(\pi^2 + (\ln x)^2)}$.

Let $x = \frac{1}{u}$, so $dx = -\frac{1}{u^2} du$. When $x=0$, $u=\infty$, and when $x=\infty$, $u=0$. Then,

$I = \int_{\infty}^{0} \frac{-\frac{1}{u^2} du}{(1+\frac{1}{u})(\pi^2 + (\ln (\frac{1}{u}))^2)} = \int_{\infty}^{0} \frac{-\frac{1}{u^2} du}{(\frac{u+1}{u})(\pi^2 + (-\ln u)^2)} = \int_{0}^{\infty} \frac{\frac{1}{u^2} du}{(\frac{u+1}{u})(\pi^2 + (\ln u)^2)} = \int_{0}^{\infty} \frac{du}{u(u+1)(\pi^2 + (\ln u)^2)} = \int_{0}^{\infty} \frac{dx}{x(x+1)(\pi^2 + (\ln x)^2)}$

$I = \int_0^{\infty} \frac{dx}{(1+x)(\pi^2 + (\ln x)^2)}$

$I = \int_0^{\infty} \frac{dx}{x(1+x)(\pi^2 + (\ln x)^2)}$

Then, adding the two equations gives

$2I = \int_0^{\infty} \frac{1}{(1+x)(\pi^2 + (\ln x)^2)} + \frac{1}{x(1+x)(\pi^2 + (\ln x)^2)} dx$

$2I = \int_0^{\infty} \frac{x+1}{x(1+x)(\pi^2 + (\ln x)^2)} dx$

$2I = \int_0^{\infty} \frac{1}{x(\pi^2 + (\ln x)^2)} dx$

Let $t = \ln x$. Then $x = e^t$, so $dx = e^t dt$. When $x=0$, $t = -\infty$. When $x=\infty$, $t=\infty$.

$2I = \int_{-\infty}^{\infty} \frac{e^t dt}{e^t (\pi^2 + t^2)} = \int_{-\infty}^{\infty} \frac{dt}{\pi^2 + t^2} = \frac{1}{\pi} \arctan \frac{t}{\pi} |_{-\infty}^{\infty} = \frac{1}{\pi} (\frac{\pi}{2} - (-\frac{\pi}{2})) = \frac{1}{\pi} (\pi) = 1$

So $2I = 1$, which means $I = \frac{1}{2}$.

3. Final Answer

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