Ответы. Интегрирование рациональных функций

Подготовка к экзамену «Элементы высшей математики» · Вопросы №12–13 · Метод неопределённых коэффициентов · Ответы к 50 задачам

Уровень А (№1–20)
  1. \(\ln|x-3| + C\)
  2. \(\ln|x+5| + C\)
  3. \(-\dfrac{1}{x+1} + C\)
  4. \(-\dfrac{1}{2(x-2)^2} + C\)
  5. \(\dfrac{1}{2}\ln|2x+1| + C\)
  6. \(\dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C\)
  7. \(\dfrac{1}{4}\ln\left|\dfrac{x-2}{x+2}\right| + C\)
  8. \(\dfrac{1}{6}\ln\left|\dfrac{x-3}{x+3}\right| + C\)
  9. \(x^2-4x+3=(x-1)(x-3)\): \(\ \dfrac{1}{2}\ln\left|\dfrac{x-3}{x-1}\right| + C\)
  10. \(x^2+5x+6=(x+2)(x+3)\): \(\ \ln\left|\dfrac{x+2}{x+3}\right| + C\)
  11. \(\dfrac{1}{2}\ln\left|\dfrac{x}{x+2}\right| + C\)
  12. \(\ln\left|\dfrac{x-1}{x}\right| + C\)
  13. \(\dfrac{1}{2}\ln\left|x^2-1\right| + C\)
  14. \(x^2-x-2=(x-2)(x+1)\): \(\ \dfrac{1}{3}\ln\left|\dfrac{x-2}{x+1}\right| + C\)
  15. \(\dfrac{1}{3}\ln\left|\dfrac{x}{x+3}\right| + C\)
  16. \(\dfrac{x+3}{(x-1)(x+1)}=\dfrac{2}{x-1}-\dfrac{1}{x+1}\): \(\ 2\ln|x-1| - \ln|x+1| + C\)
  17. \(\dfrac{2x+1}{x(x+1)}=\dfrac{1}{x}+\dfrac{1}{x+1}\): \(\ \ln|x| + \ln|x+1| + C = \ln\left|x^2+x\right| + C\)
  18. \(x^2-6x+8=(x-2)(x-4)\): \(\ \dfrac{1}{2}\ln\left|\dfrac{x-4}{x-2}\right| + C\)
  19. \(\dfrac{x-5}{(x-1)(x+1)}=\dfrac{-2}{x-1}+\dfrac{3}{x+1}\): \(\ -2\ln|x-1| + 3\ln|x+1| + C\)
  20. \(\dfrac{1}{5}\ln\left|\dfrac{x-3}{x+2}\right| + C\)
Уровень Б (№21–40)
  1. \(\dfrac{1}{x(x-1)(x+1)}=-\dfrac{1}{x}+\dfrac{1/2}{x-1}+\dfrac{1/2}{x+1}\): \(\ \dfrac{1}{2}\ln\left|x^2-1\right| - \ln|x| + C\)
  2. \(\dfrac{1/2}{x-1}-\dfrac{1}{x-2}+\dfrac{1/2}{x-3}\): \(\ \dfrac{1}{2}\ln|x-1| - \ln|x-2| + \dfrac{1}{2}\ln|x-3| + C\)
  3. \(\dfrac{2x-1}{x(x-1)(x-2)}=-\dfrac{1/2}{x}-\dfrac{1}{x-1}+\dfrac{3/2}{x-2}\): \(\ -\dfrac{1}{2}\ln|x| - \ln|x-1| + \dfrac{3}{2}\ln|x-2| + C\)
  4. \(\dfrac{1}{x^2(x+1)}=-\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x+1}\): \(\ \ln\left|\dfrac{x+1}{x}\right| - \dfrac{1}{x} + C\)
  5. \(\dfrac{x+1}{x^2(x-1)}=-\dfrac{2}{x}-\dfrac{1}{x^2}+\dfrac{2}{x-1}\): \(\ 2\ln\left|\dfrac{x-1}{x}\right| + \dfrac{1}{x} + C\)
  6. \(\dfrac{x}{(x+1)^2(x-1)}=-\dfrac{1/4}{x+1}+\dfrac{1/2}{(x+1)^2}+\dfrac{1/4}{x-1}\): \(\ \dfrac{1}{4}\ln\left|\dfrac{x-1}{x+1}\right| - \dfrac{1}{2(x+1)} + C\)
  7. \(\dfrac{3x+2}{x(x+1)^2}=\dfrac{2}{x}-\dfrac{2}{x+1}+\dfrac{1}{(x+1)^2}\): \(\ 2\ln\left|\dfrac{x}{x+1}\right| - \dfrac{1}{x+1} + C\)
  8. \(x^2+2x+5=(x+1)^2+4\): \(\ \dfrac{1}{2}\,\operatorname{arctg}\dfrac{x+1}{2} + C\)
  9. \(x^2+4x+13=(x+2)^2+9\): \(\ \dfrac{1}{3}\,\operatorname{arctg}\dfrac{x+2}{3} + C\)
  10. \(x^2-6x+10=(x-3)^2+1\): \(\ \operatorname{arctg}(x-3) + C\)
  11. \(\dfrac{1}{2}\ln\left(x^2+4x+13\right) + \dfrac{1}{3}\,\operatorname{arctg}\dfrac{x+2}{3} + C\)
  12. \(\ln\left(x^2+2x+2\right) + \operatorname{arctg}(x+1) + C\)
  13. \(\dfrac{1}{2}\ln\left(x^2+6x+10\right) - 4\,\operatorname{arctg}(x+3) + C\)
  14. \(\dfrac{x^2}{x-1}=x+1+\dfrac{1}{x-1}\): \(\ \dfrac{x^2}{2} + x + \ln|x-1| + C\)
  15. \(\dfrac{x^2+1}{x+2}=x-2+\dfrac{5}{x+2}\): \(\ \dfrac{x^2}{2} - 2x + 5\ln|x+2| + C\)
  16. \(\dfrac{x^3}{x+1}=x^2-x+1-\dfrac{1}{x+1}\): \(\ \dfrac{x^3}{3} - \dfrac{x^2}{2} + x - \ln|x+1| + C\)
  17. \(\dfrac{x^2+3x+5}{x+1}=x+2+\dfrac{3}{x+1}\): \(\ \dfrac{x^2}{2} + 2x + 3\ln|x+1| + C\)
  18. \(\dfrac{x^3}{x^2-1}=x+\dfrac{x}{x^2-1}\): \(\ \dfrac{x^2}{2} + \dfrac{1}{2}\ln\left|x^2-1\right| + C\)
  19. \(\dfrac{x^2+2}{x^2-1}=1+\dfrac{3}{x^2-1}\): \(\ x + \dfrac{3}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C\)
  20. \(\dfrac{2x^3+3x^2-1}{x^2+x}=2x+1-\dfrac{1}{x}\): \(\ x^2 + x - \ln|x| + C\)
Уровень В (№41–50)
  1. \(\dfrac{1}{(x-1)(x^2+1)}=\dfrac{1/2}{x-1}-\dfrac{1}{2}\cdot\dfrac{x+1}{x^2+1}\): \(\ \dfrac{1}{2}\ln|x-1| - \dfrac{1}{4}\ln\left(x^2+1\right) - \dfrac{1}{2}\,\operatorname{arctg} x + C\)
  2. \(x^3+8=(x+2)(x^2-2x+4)\), \(\dfrac{1}{x^3+8}=\dfrac{1/12}{x+2}+\dfrac{-x+4}{12\left(x^2-2x+4\right)}\): \(\ \dfrac{1}{12}\ln|x+2| - \dfrac{1}{24}\ln\left(x^2-2x+4\right) + \dfrac{\sqrt{3}}{12}\,\operatorname{arctg}\dfrac{x-1}{\sqrt{3}} + C\)
  3. \(\dfrac{1}{x^4-1}=\dfrac{1}{2}\left(\dfrac{1}{x^2-1}-\dfrac{1}{x^2+1}\right)\): \(\ \dfrac{1}{4}\ln\left|\dfrac{x-1}{x+1}\right| - \dfrac{1}{2}\,\operatorname{arctg} x + C\)
  4. \(\dfrac{1}{x^2\left(x^2+1\right)}=\dfrac{1}{x^2}-\dfrac{1}{x^2+1}\): \(\ -\dfrac{1}{x} - \operatorname{arctg} x + C\)
  5. \(x^3-x^2+x-1=(x-1)\left(x^2+1\right)\), \(\dfrac{x^3+1}{x^3-x^2+x-1}=1+\dfrac{1}{x-1}-\dfrac{1}{x^2+1}\): \(\ x + \ln|x-1| - \operatorname{arctg} x + C\)
  6. \(\dfrac{1}{x\left(x^2+4\right)}=\dfrac{1/4}{x}-\dfrac{x}{4\left(x^2+4\right)}\): \(\ \dfrac{1}{4}\ln|x| - \dfrac{1}{8}\ln\left(x^2+4\right) + C\)
  7. замена \(t=x-1\): \(\dfrac{x^2}{(x-1)^3}=\dfrac{1}{x-1}+\dfrac{2}{(x-1)^2}+\dfrac{1}{(x-1)^3}\): \(\ \ln|x-1| - \dfrac{2}{x-1} - \dfrac{1}{2(x-1)^2} + C\)
  8. \(x^3-1=(x-1)\left(x^2+x+1\right)\), \(\dfrac{1}{x^3-1}=\dfrac{1/3}{x-1}-\dfrac{x+2}{3\left(x^2+x+1\right)}\): \(\ \dfrac{1}{3}\ln|x-1| - \dfrac{1}{6}\ln\left(x^2+x+1\right) - \dfrac{\sqrt{3}}{3}\,\operatorname{arctg}\dfrac{2x+1}{\sqrt{3}} + C\)
  9. \(\dfrac{x+1}{x\left(x^2+1\right)}=\dfrac{1}{x}+\dfrac{-x+1}{x^2+1}\): \(\ \ln|x| - \dfrac{1}{2}\ln\left(x^2+1\right) + \operatorname{arctg} x + C\)
  10. \(\dfrac{x^4+1}{x^3-x}=x+\dfrac{x^2+1}{x(x-1)(x+1)}=x-\dfrac{1}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+1}\): \(\ \dfrac{x^2}{2} - \ln|x| + \ln\left|x^2-1\right| + C\)