Đáp án:

$\rm Ci(1)\sin(1) - Si(1)\cos(1) + \dfrac12\pi\cos(1)$

Giải thích các bước giải:

$\begin{array}{l}\quad I = \displaystyle\int\limits_0^1\dfrac{1}{\ln^2x +1}dx\\ Đặt\,\,u= \ln x\\ \to du = \dfrac{1}{x}dx\\ \text{Đổi cận:}\\ x\quad\Big|\qquad 0 \qquad\,\,\,\, 1\\ \overline{u\quad\, \Big|\quad -\infty\qquad 0}\\ \text{Ta được:}\\ \quad I = \displaystyle\int\limits_{-\infty}^0\dfrac{e^{u}}{u^2 +1}du\\ Đặt\,\,t = -u\\ \text{Ta được:}\\ \quad I = \displaystyle\int\limits_0^{+\infty}\dfrac{e^{-t}}{t^2 +1}dt\\ \text{Ta có:}\\ \quad F(s) = \displaystyle\int\limits_0^{+\infty}e^{-st}\dfrac{1}{t^2 +1}dt= \mathfrak{L}\left\{\dfrac{1}{t^2+1}\right\}\\ \to F(s) = \rm Ci(s)\sin(s) + \dfrac12(\pi - 2Si(s))\cos(s)\qquad \text{(Laplace transform)}\\ \text{Với $s=1$ ta được:}\\ \quad F(1) = I = \displaystyle\int\limits_0^{+\infty}e^{-t}\dfrac{1}{t^2 +1}dt\\ \to I = \lim\limits_{s \to 1}\left[\rm Ci(s)\sin(s) + \dfrac12(\pi - 2Si(s))\cos(s)\right]\\ \to I = \rm Ci(1)\sin(1) + \dfrac12(\pi - 2Si(1))\cos(1)\approx 0.62145\\ \text{Trong đó:}\\ \quad \begin{cases}\rm Ci(s) = -\displaystyle\int\limits_s^{+\infty}\dfrac{\cos t}{t}dt\\\rm Si(s) = \displaystyle\int\limits_0^s\dfrac{\sin t}{t}dt\end{cases} \end{array}$