$1)\displaystyle\int \dfrac{x^4-2}{x^3-x} \, dx\\ =\displaystyle\int \dfrac{x^4-x^2+\dfrac{1}{3}(3x^2-1)-\dfrac{5}{3}}{x^3-x} \, dx\\ =\displaystyle\int \left( x+\dfrac{1}{3}.\dfrac{3x^2-1}{x^3-x}-\dfrac{5}{3}.\dfrac{1}{x^3-x} \right) \, dx\\ =\displaystyle\int x \, dx+ \dfrac{1}{3}\displaystyle\int \dfrac{d(x^3-x)}{x^3-x}-\dfrac{5}{3}\displaystyle\int \dfrac{1}{x(x-1)(x+1)} \, dx\\ =\dfrac{x^2}{2}+\dfrac{1}{3}ln(|x(x-1)(x+1)|)-\dfrac{5}{3}\displaystyle\int \left(\dfrac{-1}{x}+\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)} \right)\, dx+C\\ =\dfrac{x^2}{2}+\dfrac{1}{3}ln(|x|)+\dfrac{1}{3}ln(|x-1|)+\dfrac{1}{3}ln(|x+1|)+\dfrac{5}{3}ln(|x|)-\dfrac{5}{6}ln(|x-1|)+\dfrac{5}{6}ln(|x+1|)+C\\ =\dfrac{x^2}{2}+2ln(|x|)-\dfrac{1}{2}ln(|x-1|)-\dfrac{1}{2}ln(|x+1|)+C\\ 2)E=\displaystyle\int \dfrac{x^4-x+1}{x^2+4} \, dx\\ =\displaystyle\int \dfrac{x^4+4x^2-4x^2-16-\dfrac{1}{2}.2x+17}{x^2+4} \, dx\\ =\displaystyle\int (x^2-4) \,dx-\dfrac{1}{2}\displaystyle\int \dfrac{2x}{x^2+4} \, dx+\displaystyle\int \dfrac{17}{x^2+4} \, dx\\ =\dfrac{x^3}{3}-\dfrac{1}{2}ln(|x^2+4|)+\displaystyle\int \dfrac{17}{x^2+4} \, dx+C\\ =\dfrac{x^3}{3}-\dfrac{1}{2}ln(|x^2+4|)+I+C\\ I=\displaystyle\int \dfrac{17}{x^2+4}\\ x=2\tan t \Rightarrow t=\arctan \dfrac{x}{2}\\ dx=2(\tan^2t+1) dt\\ I=17\displaystyle\int \dfrac{1}{4\tan^2t+4} 2(\tan^2t+1) dt\\ =\dfrac{17}{2}\displaystyle\int dt\\ =\dfrac{17}{2}t+C\\ =\dfrac{17}{2}\arctan \dfrac{x}{2}+C\\ =>E=\dfrac{x^3}{3}-\dfrac{1}{2}ln(|x^2+4|)+\dfrac{17}{2}\arctan \dfrac{x}{2}+C$
