$\begin{array}{l}a)\quad I= \displaystyle\int\dfrac{x^2+1}{x^4 + x^2 + 1}dx\\ \to I = \dfrac12\displaystyle\int\dfrac{1}{x^2 + x + 1}dx + \dfrac12\displaystyle\int\dfrac{1}{x^2 -x +1}dx\\ \to I = \dfrac12\displaystyle\int\dfrac{1}{\left(x + \dfrac12\right)^2 + \dfrac34}dx + \dfrac12\displaystyle\int\dfrac{1}{\left(x - \dfrac12\right)^2 + \dfrac34}dx\\ Đặt\,\,\begin{cases}u = x + \dfrac12\\v = x - \dfrac12 \end{cases}\longrightarrow \begin{cases}du = dx\\dv = dx\end{cases}\\ \to I = \dfrac23\displaystyle\int\dfrac{1}{\dfrac{4u^2}{3} + 1}du + \dfrac23\displaystyle\int\dfrac{1}{\dfrac{4v^2}{3} + 1}dv\\ Đặt\,\,\begin{cases}t = \dfrac{2u}{\sqrt3}\\z = \dfrac{2v}{\sqrt3}\end{cases}\longrightarrow \begin{cases}dt = \dfrac{2}{\sqrt3}du\\dz = \dfrac{2}{\sqrt3}du\end{cases}\\ \to I = \dfrac{1}{\sqrt3}\displaystyle\int\dfrac{1}{t^2 +1}dt + \dfrac{1}{\sqrt3}\displaystyle\int\dfrac{1}{z^2+1}dz\\ \to I = \dfrac{1}{\sqrt3}\arctan t + \dfrac{1}{\sqrt3}\arctan z + C\\ \to I = \dfrac{1}{\sqrt3}\arctan\dfrac{2u}{\sqrt3} + \dfrac{1}{\sqrt3}\arctan\dfrac{2v}{\sqrt3} + C\\ \to I = \dfrac{1}{\sqrt3}\arctan\left(\dfrac{2x+1}{\sqrt3}\right) + \dfrac{1}{\sqrt3}\arctan\left(\dfrac{2x-1}{\sqrt3}\right)+C\\ b)\quad I = \displaystyle\int\dfrac{5x-14}{x^3 - x^2 - 4x + 4}dx\\ \to I = \displaystyle\int\left(\dfrac{3}{x-1} - \dfrac{1}{x-2} - \dfrac{2}{x+2}\right)dx\\ \to I = 3\displaystyle\int\dfrac{1}{x-1}d(x-1) - \displaystyle\int\dfrac{1}{x-2}d(x-2) - 2\displaystyle\int\dfrac{1}{x+2}d(x+2)\\ \to I = 3\ln|x-1| - \ln|x-2| - 2\ln|x+2| + C\\ c)\quad I = \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\dfrac{x}{\sin^2x}dx\\ Đặt\,\,\begin{cases}u = x\\dv = \dfrac{1}{\sin^2x}dx\end{cases}\longrightarrow \begin{cases}du = dx\\v = -\cot x\end{cases}\\ \text{Ta được:}\\ \quad I = - x\cot x\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} + \displaystyle\int\limits_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\cot xdx\\ \to I = - x\cot x\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} + \ln|\sin x|\Bigg|_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}\\ \to I = \dfrac12\ln\dfrac32 - \dfrac{\pi\sqrt3}{9} + \dfrac{\pi}{4}\\ d)\quad I = \displaystyle\int\limits_0^1\dfrac{\sqrt[3]{x}}{1+x^2}dx\\ Đặt\,\,u = x^2\\ \to du = 2xdx\\ \text{Đổi cận:}\\ x \quad \Big| \quad 0 \qquad 1\\ \overline{u\quad \Big|\quad 0 \qquad 1}\\ \text{Ta được:}\\ \quad I = \dfrac12\displaystyle\int\limits_0^1\dfrac{1}{\sqrt[3]{u}(u+1)}du\\ Đặt\,\,t = \sqrt[3]{u}\\ \to dt = \dfrac{1}{\sqrt[3]{u^2}}du\\ \text{Đổi cận:}\\ u \quad \Big| \quad 0 \qquad 1\\ \overline{t\quad \Big|\quad 0 \qquad 1}\\ \text{Ta được:}\\ \quad I = \dfrac12\displaystyle\int\limits_0^1\dfrac{3t}{t^3 +1}dt\\ \to I = \dfrac12\displaystyle\int\limits_0^1\left(\dfrac{2t-1}{2(t^2 - t + 1)} + \dfrac{3}{2(t^2 - t + 1)}- \dfrac{1}{t+1}\right)dt\\ \to I = \dfrac14\displaystyle\int\limits_0^1\dfrac{d(t^2 - t)}{t^2 - t + 1} + \dfrac34\displaystyle\int\limits_0^1\dfrac{1}{t^2 - t + 1}dt - \dfrac12\displaystyle\int\limits_0^1\dfrac{1}{t+1}dt\\ \to I = \dfrac14\ln|t^2 - t + 1|\Bigg|_0^1 + \dfrac{\sqrt3}{2}\arctan\left(\dfrac{2t -1}{\sqrt3} \right)\Bigg|_0^1-\dfrac12\ln|t+1|\Bigg|_0^1\\ \to I = \dfrac{\pi\sqrt3}{6} - \dfrac12\ln2 \end{array}$
