Đáp án:
$\begin{array}{l}4)\quad I=\dfrac{\pi}{6}\\5)\quad I =\dfrac{\pi}{4}\\6)\quad I =\dfrac{\pi}{2}\end{array}$
Giải thích các bước giải:
$\begin{array}{l}4)\quad I =\displaystyle\int\limits_0^{+\infty}\dfrac{x^2}{x^6+1}dx\\\to I = \lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}\dfrac{x^2}{x^6+1}dx\\\to I = \dfrac13\lim\limits_{t \to +\infty}\displaystyle\int\limits_0^{t}\dfrac{d(x^3)}{(x^3)^2+1}\\\to I =\dfrac13\lim\limits_{t\to+\infty}\arctan x^3\Bigg|_0^t\\\to I = \dfrac13\cdot\left(\arctan(+\infty) - \arctan0\right)\\\to I =\dfrac{\pi}{6}\\5)\quad I = \displaystyle\int\limits_0^{+\infty}\dfrac{dx}{e^x + e^{-x}}\\\to I = \lim\limits_{t\to +\infty}\displaystyle\int\limits_0^t\dfrac{dx}{e^x + e^{-x}}\\\to I=\lim\limits_{t\to +\infty}\displaystyle\int\limits_0^t\dfrac{e^x}{e^{2x} + 1}dx\\\to I=\lim\limits_{t\to +\infty}\displaystyle\int\limits_0^t\dfrac{d(e^x)}{\left(e^{x}\right)^2 + 1}\\\to I =\lim\limits_{t\to +\infty}\arctan e^x\Bigg|_0^t\\\to I = \arctan e^{+\infty} - \arctan e^0\\\to I = \dfrac{\pi}{4}\\6)\quad I = \displaystyle\int\limits_1^{+\infty}\dfrac{dx}{x(\ln^2x+1)}\\\to I = \lim\limits_{t\to +\infty}\displaystyle\int\limits_1^t\dfrac{dx}{x(\ln^2x+1)}\\\to I = \lim\limits_{t\to +\infty}\displaystyle\int\limits_1^t\dfrac{d(\ln x)}{\ln^2x +1}\\\to I =\lim\limits_{t \to +\infty}\arctan(\ln x)\Bigg|_1^t\\\to I = \arctan[\ln(+\infty)] - \arctan(\ln1)\\\to I = \dfrac{\pi}{2}\end{array}$
