ព្រីមីទីវ និងអាំងតេក្រាលមិនកំណត់

មេរៀនទី ១ ៖ ព្រីមីទីវ និងអាំងតេក្រាលមិនកំណត់

ខ. ដំណោះស្រាយលម្អិតនៃលំហាត់អនុវត្តន៍

ដំណោះស្រាយ

1. \(\displaystyle I = \int 10x \sqrt{5x^2+2}\,dx\)

   តាង \(u = 5x^2+2 \implies du = 10x\,dx\)

   គេបាន \(I = \displaystyle\int \sqrt{u}\,du = \int u^{1/2}du = \frac{2}{3}u^{3/2} + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{3}(5x^2+2)\sqrt{5x^2+2} + C}\)

2. \(\displaystyle I = \int \frac{x^2}{(x^3+4)^2}\,dx\)

   តាង \(u = x^3+4 \implies du = 3x^2\,dx \implies x^2\,dx = \frac{1}{3}du\)

   គេបាន \(I = \displaystyle\frac{1}{3}\int \frac{du}{u^2} = -\frac{1}{3u} + C\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{3(x^3+4)} + C}\)

3. \(\displaystyle I = \int -6x(1-2x^2)^3\,dx\)

   តាង \(u = 1-2x^2 \implies du = -4x\,dx \implies -6x\,dx = \frac{3}{2}du\)

   គេបាន \(I = \displaystyle\frac{3}{2}\int u^3\,du = \frac{3}{8}u^4 + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{3}{8}(1-2x^2)^4 + C}\)

4. \(\displaystyle I = \int 3\sqrt{x}(9+2\sqrt{x^3})^4\,dx\)

   សម្គាល់ \(2\sqrt{x^3} = 2x^{3/2}\) នាំឱ្យ \((9+2x^{3/2})' = 3x^{1/2} = 3\sqrt{x}\)

   តាង \(u = 9+2x^{3/2} \implies du = 3\sqrt{x}\,dx\)

   គេបាន \(I = \displaystyle\int u^4\,du = \frac{1}{5}u^5 + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{5}(9+2x\sqrt{x})^5 + C}\)

5. \(\displaystyle I = \int \frac{2x+3}{(x^2+3x+1)^3}\,dx\)

   តាង \(u = x^2+3x+1 \implies du = (2x+3)\,dx\)

   គេបាន \(I = \displaystyle\int \frac{du}{u^3} = \int u^{-3}du = -\frac{1}{2u^2} + C\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{2(x^2+3x+1)^2} + C}\)

6. \(\displaystyle I = \int \frac{4x}{\sqrt{x^2+2}}\,dx\)

   តាង \(u = x^2+2 \implies du = 2x\,dx \implies 4x\,dx = 2du\)

   គេបាន \(I = \displaystyle 2\int \frac{du}{\sqrt{u}} = 4\sqrt{u} + C\)

   ដូចនេះ៖ \(\boxed{I = 4\sqrt{x^2+2} + C}\)

7. \(\displaystyle I = \int \frac{x^2+1}{\sqrt{x^3+3x+1}}\,dx\)

   តាង \(u = x^3+3x+1 \implies du = 3(x^2+1)\,dx \implies (x^2+1)\,dx = \frac{1}{3}du\)

   គេបាន \(I = \displaystyle\frac{1}{3}\int \frac{du}{\sqrt{u}} = \frac{2}{3}\sqrt{u} + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{3}\sqrt{x^3+3x+1} + C}\)

8. \(\displaystyle I = \int x\sqrt{3+4x^2}\,dx\)

   តាង \(u = 3+4x^2 \implies du = 8x\,dx \implies x\,dx = \frac{1}{8}du\)

   គេបាន \(I = \displaystyle\frac{1}{8}\int u^{1/2}du = \frac{1}{8}\cdot\frac{2}{3}u^{3/2} + C = \frac{1}{12}u\sqrt{u} + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{12}(3+4x^2)\sqrt{3+4x^2} + C}\)

9. \(\displaystyle I = \int \frac{x+1}{(x^2+2x-1)^2}\,dx\)

   តាង \(u = x^2+2x-1 \implies du = 2(x+1)\,dx \implies (x+1)\,dx = \frac{1}{2}du\)

   គេបាន \(I = \displaystyle\frac{1}{2}\int \frac{du}{u^2} = -\frac{1}{2u} + C\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{2(x^2+2x-1)} + C}\)

10. \(\displaystyle I = \int \frac{x^3}{(1-x^4)^4}\,dx\)

    តាង \(u = 1-x^4 \implies du = -4x^3\,dx \implies x^3\,dx = -\frac{1}{4}du\)

    គេបាន \(I = \displaystyle -\frac{1}{4}\int u^{-4}du = \frac{1}{12u^3} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{12(1-x^4)^3} + C}\)

11. \(\displaystyle I = \int 3e^{3x}\sqrt{1+e^{3x}}\,dx\)

    តាង \(u = 1+e^{3x} \implies du = 3e^{3x}\,dx\)

    គេបាន \(I = \displaystyle\int \sqrt{u}\,du = \frac{2}{3}u^{3/2} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{2}{3}(1+e^{3x})\sqrt{1+e^{3x}} + C}\)

12. \(\displaystyle I = \int \frac{e^x}{\sqrt{e^x+2}}\,dx\)

    តាង \(u = e^x+2 \implies du = e^x\,dx\)

    គេបាន \(I = \displaystyle\int \frac{du}{\sqrt{u}} = 2\sqrt{u} + C\)

    ដូចនេះ៖ \(\boxed{I = 2\sqrt{e^x+2} + C}\)

13. \(\displaystyle I = \int \frac{e^{-2x}}{(1-e^{-2x})^3}\,dx\)

    តាង \(u = 1-e^{-2x} \implies du = 2e^{-2x}\,dx \implies e^{-2x}\,dx = \frac{1}{2}du\)

    គេបាន \(I = \displaystyle\frac{1}{2}\int u^{-3}du = -\frac{1}{4u^2} + C\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{4(1-e^{-2x})^2} + C}\)

14. \(\displaystyle I = \int 10x^2(x^3+1)^4\,dx\)

    តាង \(u = x^3+1 \implies du = 3x^2\,dx \implies 10x^2\,dx = \frac{10}{3}du\)

    គេបាន \(I = \displaystyle\frac{10}{3}\int u^4\,du = \frac{2}{3}u^5 + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{2}{3}(x^3+1)^5 + C}\)

15. \(\displaystyle I = \int (x^3+2)(x^4+8x-3)^5\,dx\)

    តាង \(u = x^4+8x-3 \implies du = 4(x^3+2)\,dx \implies (x^3+2)\,dx = \frac{1}{4}du\)

    គេបាន \(I = \displaystyle\frac{1}{4}\int u^5\,du = \frac{1}{24}u^6 + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{24}(x^4+8x-3)^6 + C}\)

16. \(\displaystyle I = \int 2x^3\sqrt{(x^4+5)^3}\,dx\)

    តាង \(u = x^4+5 \implies du = 4x^3\,dx \implies 2x^3\,dx = \frac{1}{2}du\)

    គេបាន \(I = \displaystyle\frac{1}{2}\int u^{3/2}du = \frac{1}{5}(x^4+5)^2\sqrt{x^4+5} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{5}(x^4+5)^2\sqrt{x^4+5} + C}\)

17. \(\displaystyle I = \int \frac{\cos 2x}{\sqrt{1-\sin 2x}}\,dx\)

    តាង \(u = 1-\sin 2x \implies du = -2\cos 2x\,dx \implies \cos 2x\,dx = -\frac{1}{2}du\)

    គេបាន \(I = \displaystyle -\frac{1}{2}\int \frac{du}{\sqrt{u}} = -\sqrt{u} + C\)

    ដូចនេះ៖ \(\boxed{I = -\sqrt{1-\sin 2x} + C}\)

18. \(\displaystyle I = \int \cos x \sqrt{4+3\sin x}\,dx\)

    តាង \(u = 4+3\sin x \implies du = 3\cos x\,dx \implies \cos x\,dx = \frac{1}{3}du\)

    គេបាន \(I = \displaystyle\frac{1}{3}\int u^{1/2}du = \frac{2}{9}u^{3/2} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{2}{9}(4+3\sin x)\sqrt{4+3\sin x} + C}\)

19. \(\displaystyle I = \int \frac{1-\sin x}{\sqrt[3]{x+\cos x}}\,dx\)

    តាង \(u = x+\cos x \implies du = (1-\sin x)\,dx\)

    គេបាន \(I = \displaystyle\int u^{-1/3}du = \frac{3}{2}u^{2/3} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{3}{2}\sqrt[3]{(x+\cos x)^2} + C}\)

20. \(\displaystyle I = \int \sin\frac{1}{2}x \cos^3\frac{1}{2}x\,dx\)

    តាង \(u = \cos\frac{1}{2}x \implies du = -\frac{1}{2}\sin\frac{1}{2}x\,dx \implies \sin\frac{1}{2}x\,dx = -2du\)

    គេបាន \(I = -2\displaystyle\int u^3\,du = -\frac{1}{2}u^4 + C\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{2}\cos^4\left(\frac{1}{2}x\right) + C}\)

21. \(\displaystyle I = \int \frac{x\cos x+\sin x}{\sqrt{1+x\sin x}}\,dx\)

    ដោយ \((1+x\sin x)' = \sin x + x\cos x\) នាំឱ្យតាង \(u = 1+x\sin x \implies du = (x\cos x+\sin x)\,dx\)

    គេបាន \(I = \displaystyle\int \frac{du}{\sqrt{u}} = 2\sqrt{u} + C\)

    ដូចនេះ៖ \(\boxed{I = 2\sqrt{1+x\sin x} + C}\)

22. \(\displaystyle I = \int \frac{\sqrt{\sqrt{x}+1}}{\sqrt{x}}\,dx\)

    តាង \(u = \sqrt{x}+1 \implies du = \frac{1}{2\sqrt{x}}\,dx \implies \frac{dx}{\sqrt{x}} = 2du\)

    គេបាន \(I = 2\displaystyle\int \sqrt{u}\,du = \frac{4}{3}(\sqrt{x}+1)\sqrt{\sqrt{x}+1} + C\)

    ដូចនេះ៖ \(\boxed{I = \frac{4}{3}(\sqrt{x}+1)\sqrt{\sqrt{x}+1} + C}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int \sin 2x\cos 3x\,dx\)

   ប្រើរូបមន្ត \(\sin a\cos b = \frac{1}{2}[\sin(a+b)+\sin(a-b)]\)

   \(I = \displaystyle\frac{1}{2}\int [\sin 5x - \sin x]dx\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{10}\cos 5x + \frac{1}{2}\cos x + C}\)

2. \(\displaystyle I = \int \sin 4x\cos 6x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\sin 10x - \sin 2x]dx\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{20}\cos 10x + \frac{1}{4}\cos 2x + C}\)

3. \(\displaystyle I = \int \sin 7x\cos 5x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\sin 12x + \sin 2x]dx\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{24}\cos 12x - \frac{1}{4}\cos 2x + C}\)

4. \(\displaystyle I = \int \sin 9x\cos 4x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\sin 13x + \sin 5x]dx\)

   ដូចនេះ៖ \(\boxed{I = -\frac{1}{26}\cos 13x - \frac{1}{10}\cos 5x + C}\)

5. \(\displaystyle I = \int \cos 2x\cos x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\cos 3x + \cos x]dx\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{6}\sin 3x + \frac{1}{2}\sin x + C}\)

6. \(\displaystyle I = \int \cos 3x\cos 5x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\cos 8x + \cos 2x]dx\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{16}\sin 8x + \frac{1}{4}\sin 2x + C}\)

7. \(\displaystyle I = \int \cos 7x\cos 3x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\cos 10x + \cos 4x]dx\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{20}\sin 10x + \frac{1}{8}\sin 4x + C}\)

8. \(\displaystyle I = \int \sin 5x\sin 8x\,dx\)

   ប្រើរូបមន្ត \(\sin a\sin b = \frac{1}{2}[\cos(a-b)-\cos(a+b)]\)

   \(I = \displaystyle\frac{1}{2}\int [\cos 3x - \cos 13x]dx\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{6}\sin 3x - \frac{1}{26}\sin 13x + C}\)

9. \(\displaystyle I = \int \cos 8x\cos 10x\,dx\)

   \(I = \displaystyle\frac{1}{2}\int [\cos 18x + \cos 2x]dx\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{36}\sin 18x + \frac{1}{4}\sin 2x + C}\)

10. \(\displaystyle I = \int \sin^2 x\cos^3 x\,dx\)

    សរសេរ \(\cos^3 x = (1-\sin^2 x)\cos x\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^2(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{3}\sin^3 x - \frac{1}{5}\sin^5 x + C}\)

11. \(\displaystyle I = \int \sin 6x\sin 2x\,dx\)

    \(I = \displaystyle\frac{1}{2}\int [\cos 4x - \cos 8x]dx\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{8}\sin 4x - \frac{1}{16}\sin 8x + C}\)

12. \(\displaystyle I = \int \sin^4 x\cos^3 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^4(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{5}\sin^5 x - \frac{1}{7}\sin^7 x + C}\)

13. \(\displaystyle I = \int \sin^6 x\cos^5 x\,dx\)

    សរសេរ \(\cos^5 x = (1-\sin^2 x)^2\cos x\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^6(1-2u^2+u^4)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{7}\sin^7 x - \frac{2}{9}\sin^9 x + \frac{1}{11}\sin^{11} x + C}\)

14. \(\displaystyle I = \int \cos^4 x\cos^3 x\,dx = \int \cos^7 x\,dx\)

    សរសេរ \(\cos^7 x = (1-\sin^2 x)^3\cos x\)

    តាង \(u = \sin x \implies I = \displaystyle\int (1-3u^2+3u^4-u^6)du\)

    ដូចនេះ៖

    \(\boxed{I = \sin x - \sin^3 x + \frac{3}{5}\sin^5 x - \frac{1}{7}\sin^7 x + C}\)

15. \(\displaystyle I = \int \sin^4 x\cos^5 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^4(1-2u^2+u^4)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{5}\sin^5 x - \frac{2}{7}\sin^7 x + \frac{1}{9}\sin^9 x + C}\)

16. \(\displaystyle I = \int \sin^8 x\cos^5 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^8(1-2u^2+u^4)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{9}\sin^9 x - \frac{2}{11}\sin^{11} x + \frac{1}{13}\sin^{13} x + C}\)

17. \(\displaystyle I = \int \cos^6 x\sin^5 x\,dx\)

    សរសេរ \(\sin^5 x = (1-\cos^2 x)^2\sin x\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^6(1-2u^2+u^4)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{7}\cos^7 x + \frac{2}{9}\cos^9 x - \frac{1}{11}\cos^{11} x + C}\)

18. \(\displaystyle I = \int \sin^5 x\cos^4 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^4(1-2u^2+u^4)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + C}\)

19. \(\displaystyle I = \int \cos^2 x\sin^3 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^2(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{3}\cos^3 x + \frac{1}{5}\cos^5 x + C}\)

20. \(\displaystyle I = \int \cos^8 x\sin^5 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^8(1-2u^2+u^4)du\)

    ដូចនេះ៖

    \(\boxed{I = -\frac{1}{9}\cos^9 x + \frac{2}{11}\cos^{11} x - \frac{1}{13}\cos^{13} x + C}\)

21. \(\displaystyle I = \int \sin^3 x\cos^6 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^6(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{7}\cos^7 x + \frac{1}{9}\cos^9 x + C}\)

22. \(\displaystyle I = \int \cos^3 x\sin^6 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^6(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{7}\sin^7 x - \frac{1}{9}\sin^9 x + C}\)

23. \(\displaystyle I = \int \sin^5 x\cos^7 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^5(1-u^2)^3du\)

    ដូចនេះ៖

    \(\boxed{I = \frac{1}{6}\sin^6 x - \frac{3}{8}\sin^8 x + \frac{3}{10}\sin^{10} x - \frac{1}{12}\sin^{12} x + C}\)

24. \(\displaystyle I = \int \sin^2 x\cos^2 x\,dx\)

    \(I = \displaystyle\frac{1}{4}\int \sin^2 2x\,dx = \frac{1}{8}\int (1-\cos 4x)dx\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{8}x - \frac{1}{32}\sin 4x + C}\)

25. \(\displaystyle I = \int \sin^3 x\cos^5 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^5(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{6}\cos^6 x + \frac{1}{8}\cos^8 x + C}\)

26. \(\displaystyle I = \int \sin^3 x\cos^7 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int u^7(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{8}\cos^8 x + \frac{1}{10}\cos^{10} x + C}\)

27. \(\displaystyle I = \int \cos^2 x\sin^4 x\,dx\)

    \(I = \displaystyle\frac{1}{8}\int (1-\cos 2x)\sin^2 2x\,dx\)

    \(= \displaystyle\frac{1}{8}\int \left(\frac{1-\cos 4x}{2} - \sin^2 2x\cos 2x\right)dx\)

    ដូចនេះ៖

    \(\boxed{I = \frac{1}{16}x - \frac{1}{64}\sin 4x - \frac{1}{48}\sin^3 2x + C}\)

28. \(\displaystyle I = \int \sin^3 x\cos^7 x\,dx\)

    តាង \(u = \cos x \implies I = -\displaystyle\int (u^7-u^9)du\)

    ដូចនេះ៖ \(\boxed{I = -\frac{1}{8}\cos^8 x + \frac{1}{10}\cos^{10} x + C}\)

29. \(\displaystyle I = \int \cos^3 x\sin^7 x\,dx\)

    តាង \(u = \sin x \implies I = \displaystyle\int u^7(1-u^2)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{8}\sin^8 x - \frac{1}{10}\sin^{10} x + C}\)

30. \(\displaystyle I = \int \tan^2 x\,dx\)

    សរសេរ \(\tan^2 x = (\tan^2 x+1) - 1\)

    ដូចនេះ៖ \(\boxed{I = \tan x - x + C}\)

31. \(\displaystyle I = \int \tan^3 x\,dx\)

    \(I = \displaystyle\int \tan x(\tan^2 x+1)dx - \int \tan x\,dx\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{2}\tan^2 x + \ln|\cos x| + C}\)

32. \(\displaystyle I = \int \cot^2 x\,dx\)

    សរសេរ \(\cot^2 x = (\cot^2 x+1) - 1\)

    ដូចនេះ៖ \(\boxed{I = -\cot x - x + C}\)

33. \(\displaystyle I = \int \tan^4 x\,dx\)

    \(I = \displaystyle\int \tan^2 x(\tan^2 x+1)dx - \int (\tan^2 x+1-1)dx\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{3}\tan^3 x - \tan x + x + C}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int \frac{4x-3}{x+2}\,dx\)

   ចែកភាគយកនឹងភាគបែង៖ \(\displaystyle\frac{4x-3}{x+2} = 4 - \frac{11}{x+2}\)

   ដូចនេះ៖ \(\boxed{I = 4x - 11\ln|x+2| + C}\)

2. \(\displaystyle I = \int \frac{3x+2}{x-4}\,dx\)

   \(\displaystyle\frac{3x+2}{x-4} = 3 + \frac{14}{x-4}\)

   ដូចនេះ៖ \(\boxed{I = 3x + 14\ln|x-4| + C}\)

3. \(\displaystyle I = \int \frac{x+1}{3x-2}\,dx\)

   \(\displaystyle\frac{x+1}{3x-2} = \frac{1}{3} + \frac{5/3}{3x-2}\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{3}x + \frac{5}{9}\ln|3x-2| + C}\)

4. \(\displaystyle I = \int \frac{5x-3}{2x+1}\,dx\)

   \(\displaystyle\frac{5x-3}{2x+1} = \frac{5}{2} - \frac{11/2}{2x+1}\)

   ដូចនេះ៖ \(\boxed{I = \frac{5}{2}x - \frac{11}{4}\ln|2x+1| + C}\)

5. \(\displaystyle I = \int \frac{2x+3}{(x-3)^2}\,dx\)

   តាង \(u = x-3 \implies x = u+3 \implies 2x+3 = 2u+9\)

   \(I = \displaystyle\int \left(\frac{2}{u} + \frac{9}{u^2}\right)du = 2\ln|u| - \frac{9}{u} + C\)

   ដូចនេះ៖ \(\boxed{I = 2\ln|x-3| - \frac{9}{x-3} + C}\)

6. \(\displaystyle I = \int \frac{x-3}{(3x-2)^2}\,dx\)

   តាង \(u = 3x-2 \implies x = \frac{u+2}{3}, dx = \frac{du}{3} \implies x-3 = \frac{u-7}{3}\)

   \(I = \displaystyle\frac{1}{9}\int \left(\frac{1}{u} - \frac{7}{u^2}\right)du = \frac{1}{9}\ln|3x-2| + \frac{7}{9(3x-2)} + C\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{9}\ln|3x-2| + \frac{7}{9(3x-2)} + C}\)

7. \(\displaystyle I = \int \frac{2x-5}{(3x-2)^2}\,dx\)

   តាង \(u = 3x-2 \implies 2x-5 = \frac{2u-11}{3}\)

   \(I = \displaystyle\frac{1}{9}\int \left(\frac{2}{u} - \frac{11}{u^2}\right)du\)

   ដូចនេះ៖ \(\boxed{I = \frac{2}{9}\ln|3x-2| + \frac{11}{9(3x-2)} + C}\)

8. \(\displaystyle I = \int \frac{x}{(2x-1)^2}\,dx\)

   តាង \(u = 2x-1 \implies x = \frac{u+1}{2}, dx = \frac{du}{2}\)

   \(I = \displaystyle\frac{1}{4}\int \left(\frac{1}{u} + \frac{1}{u^2}\right)du\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{4}\ln|2x-1| - \frac{1}{4(2x-1)} + C}\)

9. \(\displaystyle I = \int \frac{x^2+x+1}{x+2}\,dx\)

   ចែកពហុធា៖ \(\displaystyle\frac{x^2+x+1}{x+2} = x - 1 + \frac{3}{x+2}\)

   ដូចនេះ៖ \(\boxed{I = \frac{1}{2}x^2 - x + 3\ln|x+2| + C}\)

10. \(\displaystyle I = \int \frac{x^2+2}{3x+2}\,dx\)

    \(\displaystyle\frac{x^2+2}{3x+2} = \frac{1}{3}x - \frac{2}{9} + \frac{22/9}{3x+2}\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{6}x^2 - \frac{2}{9}x + \frac{22}{27}\ln|3x+2| + C}\)

11. \(\displaystyle I = \int \frac{x^2-3x+5}{2x+1}\,dx\)

    \(\displaystyle\frac{x^2-3x+5}{2x+1} = \frac{1}{2}x - \frac{7}{4} + \frac{27/4}{2x+1}\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{4}x^2 - \frac{7}{4}x + \frac{27}{8}\ln|2x+1| + C}\)

12. \(\displaystyle I = \int \frac{2x^2-x+3}{x-3}\,dx\)

    \(\displaystyle\frac{2x^2-x+3}{x-3} = 2x + 5 + \frac{18}{x-3}\)

    ដូចនេះ៖ \(\boxed{I = x^2 + 5x + 18\ln|x-3| + C}\)

13. \(\displaystyle I = \int \frac{2x^2+x+1}{(x+2)^2}\,dx\)

    តាង \(u = x+2 \implies 2(u-2)^2+(u-2)+1 = 2u^2-7u+7\)

    \(I = \displaystyle\int \left(2 - \frac{7}{u} + \frac{7}{u^2}\right)du\)

    ដូចនេះ៖ \(\boxed{I = 2x - 7\ln|x+2| - \frac{7}{x+2} + C}\)

14. \(\displaystyle I = \int \frac{x^2-3x+4}{(2x+3)^2}\,dx\)

    តាង \(u = 2x+3 \implies x^2-3x+4 = \frac{u^2-12u+31}{4}\)

    \(I = \displaystyle\frac{1}{8}\int \left(1 - \frac{12}{u} + \frac{31}{u^2}\right)du\)

    ដូចនេះ៖ \(\boxed{I = \frac{1}{4}x - \frac{3}{2}\ln|2x+3| - \frac{31}{8(2x+3)} + C}\)

15. \(\displaystyle I = \int \frac{x^2+1}{(x-1)^2}\,dx\)

    តាង \(u = x-1 \implies (u+1)^2+1 = u^2+2u+2\)

    \(I = \displaystyle\int \left(1 + \frac{2}{u} + \frac{2}{u^2}\right)du\)

    ដូចនេះ៖ \(\boxed{I = x + 2\ln|x-1| - \frac{2}{x-1} + C}\)

16. \(\displaystyle I = \int \frac{x^2}{x^2-5x+6}\,dx\)

    បំបែកជាប្រភាគងាយ៖ \(\displaystyle\frac{x^2}{x^2-5x+6} = 1 - \frac{4}{x-2} + \frac{9}{x-3}\)

    ដូចនេះ៖ \(\boxed{I = x - 4\ln|x-2| + 9\ln|x-3| + C}\)

17. \(\displaystyle I = \int \frac{x^2+4}{x^2-4}\,dx\)

    \(\displaystyle\frac{x^2+4}{x^2-4} = 1 + \frac{2}{x-2} - \frac{2}{x+2}\)

    ដូចនេះ៖ \(\boxed{I = x + 2\ln\left|\frac{x-2}{x+2}\right| + C}\)

18. \(\displaystyle I = \int \frac{x^2+x+1}{(x+1)(x+2)}\,dx\)

    \(\displaystyle\frac{x^2+x+1}{(x+1)(x+2)} = 1 + \frac{1}{x+1} - \frac{3}{x+2}\)

    ដូចនេះ៖ \(\boxed{I = x + \ln|x+1| - 3\ln|x+2| + C}\)

19. \(\displaystyle I = \int \frac{x+1}{(x+2)(x+3)}\,dx\)

    បំបែកជាប្រភាគងាយ៖ \(\displaystyle\frac{x+1}{(x+2)(x+3)} = -\frac{1}{x+2} + \frac{2}{x+3}\)

    ដូចនេះ៖ \(\boxed{I = -\ln|x+2| + 2\ln|x+3| + C}\)

20. \(\displaystyle I = \int \frac{2x+3}{(x-1)(x-2)}\,dx\)

    \(\displaystyle\frac{2x+3}{(x-1)(x-2)} = -\frac{5}{x-1} + \frac{7}{x-2}\)

    ដូចនេះ៖ \(\boxed{I = -5\ln|x-1| + 7\ln|x-2| + C}\)

21. \(\displaystyle I = \int \frac{3}{(2x+3)(3x+4)}\,dx\)

    \(\displaystyle\frac{3}{(2x+3)(3x+4)} = -\frac{6}{2x+3} + \frac{9}{3x+4}\)

    ដូចនេះ៖ \(\boxed{I = 3\ln\left|\frac{3x+4}{2x+3}\right| + C}\)

ដំណោះស្រាយ

1. សរសេរ \(f(x)\) ជារាង \(f(x)=\dfrac{A}{x-1}+\dfrac{B}{x-2}+\dfrac{C}{x-4}\)

   គុណអង្គទាំងពីរនឹង \((x-1)(x-2)(x-4)\) គេបាន៖

   \[\begin{aligned} x^2+2x+6 &= A(x-2)(x-4) \\ &\quad{} + B(x-1)(x-4) \\ &\quad{} + C(x-1)(x-2) \end{aligned}\]

   ជំនួស \(x = 1\)៖ \(1^2+2(1)+6 = A(1-2)(1-4) \implies 9 = 3A \implies A = 3\)

   ជំនួស \(x = 2\)៖ \(2^2+2(2)+6 = B(2-1)(2-4) \implies 14 = -2B \implies B = -7\)

   ជំនួស \(x = 4\)៖ \(4^2+2(4)+6 = C(4-1)(4-2) \implies 30 = 6C \implies C = 5\)

   ដូចនេះ៖ \(\boxed{A = 3,\; B = -7,\; C = 5}\) ហើយគេសរសេរបាន៖

   \[\boxed{f(x) = \frac{3}{x-1} - \frac{7}{x-2} + \frac{5}{x-4}}\]

2. គណនាអាំងតេក្រាលមិនកំណត់ \(\displaystyle\int f(x) dx\)

   ធ្វើអាំងតេក្រាលលើកន្សោមប្រភាគងាយ៖

   \[\begin{aligned} I &= \int \left(\frac{3}{x-1} - \frac{7}{x-2} + \frac{5}{x-4}\right)dx \\ &= 3\ln|x-1| - 7\ln|x-2| \\ &\quad{} + 5\ln|x-4| + C \end{aligned}\]

   ដូចនេះ៖

   \(\boxed{I = 3\ln|x-1| - 7\ln|x-2| + 5\ln|x-4| + C}\)

ដំណោះស្រាយ

1. កំណត់តម្លៃ \(A, B, C\) និង \(D\)

   តម្រូវភាគបែងរួម គេបាន៖

   \[\begin{aligned} x^2+1 &= A(x+3) + B(x-1)(x+3) \\ &\quad{} + C(x-1)^2(x+3) + D(x-1)^3 \end{aligned}\]

   យក \(x = 1\)៖ \(1^2+1 = A(1+3) \implies 4A = 2 \implies A = \dfrac{1}{2}\)

   យក \(x = -3\)៖ \((-3)^2+1 = D(-3-1)^3 \implies -64D = 10 \implies D = -\dfrac{5}{32}\)

   ផ្ទឹមមេគុណនៃ \(x^3\)៖ \(C + D = 0 \implies C = -D = \dfrac{5}{32}\)

   យក \(x = 0\)៖ \(1 = 3A - 3B + 3C - D \implies 3B = \frac{9}{8} \implies B = \dfrac{3}{8}\)

   ដូចនេះ៖ \(\boxed{A = \frac{1}{2},\; B = \frac{3}{8},\; C = \frac{5}{32},\; D = -\frac{5}{32}}\)

2. គណនាអាំងតេក្រាលមិនកំណត់ \(\displaystyle\int f(x) dx\)

   ធ្វើអាំងតេក្រាលលើកន្សោមប្រភាគនីមួយៗ៖

   \[\begin{aligned} I &= \int \left[\frac{1/2}{(x-1)^3} + \frac{3/8}{(x-1)^2}\right]dx \\ &\quad{} + \frac{5}{32}\int \left(\frac{1}{x-1} - \frac{1}{x+3}\right)dx \\ &= -\frac{1}{4(x-1)^2} - \frac{3}{8(x-1)} \\ &\quad{} + \frac{5}{32}\ln\left|\frac{x-1}{x+3}\right| + C \end{aligned}\]

   ដូចនេះ៖

   \(\boxed{I = -\frac{1}{4(x-1)^2} - \frac{3}{8(x-1)} + \frac{5}{32}\ln\left|\frac{x-1}{x+3}\right| + C}\)

ដំណោះស្រាយ

1. សរសេរ \(f(x)\) ជារាង \(f(x)=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^{2}}\)

   ដាក់ភាគបែងជាផលគុណកត្តា៖ \(x^3+2x^2+x = x(x+1)^2\)

   តម្រូវភាគបែងរួម៖ \(5x^2+20x+6 = A(x+1)^2 + Bx(x+1) + Cx\)

   យក \(x = 0\)៖ \(6 = A(1)^2 \implies A = 6\)

   យក \(x = -1\)៖ \(5(-1)^2+20(-1)+6 = C(-1) \implies -9 = -C \implies C = 9\)

   ផ្ទឹមមេគុណនៃ \(x^2\)៖ \(5 = A + B \implies 5 = 6 + B \implies B = -1\)

   ដូចនេះ៖ \(\boxed{A = 6,\; B = -1,\; C = 9}\) ហើយគេបាន \(\boxed{f(x) = \frac{6}{x} - \frac{1}{x+1} + \frac{9}{(x+1)^2}}\)

2. គណនាអាំងតេក្រាលមិនកំណត់ \(\displaystyle\int f(x) dx\)

   ធ្វើអាំងតេក្រាល៖

   \[\begin{aligned} \int f(x)dx &= \int \left(\frac{6}{x} - \frac{1}{x+1} + \frac{9}{(x+1)^2}\right)dx \\ &= 6\ln|x| - \ln|x+1| - \frac{9}{x+1} + C \end{aligned}\]

   ដូចនេះ៖ \(\boxed{I = 6\ln|x| - \ln|x+1| - \frac{9}{x+1} + C}\)

ដំណោះស្រាយ

1. កំណត់តម្លៃ \(A, B\)

   តម្រូវភាគបែងរួម គេបាន
   

   \[\begin{aligned} f(x) &= \frac{A(x-1)^2 + B(x+2)}{(x+2)(x-1)^2} \\ &= \frac{Ax^2 + (-2A+B)x + (A+2B)}{(x+2)(x-1)^2} \end{aligned}\]

   ផ្ទឹមនឹងភាគយក \(3x^2-10x-5\) គេបាន៖

   \[\begin{cases} A = 3 \\ -2A + B = -10 \implies B = -4 \\ A + 2B = 3 - 8 = -5 \quad{} (\text{ផ្ទៀងផ្ទាត់}) \end{cases}\]

   ដូចនេះ៖ \(\boxed{A = 3,\; B = -4}\)

   គេបាន៖ \(\boxed{f(x) = \frac{3}{x+2} - \frac{4}{(x-1)^2}}\)

2. គណនាអាំងតេក្រាល \(I(x)=\displaystyle\int f(x) dx\) ដោយដឹងថា \(I(0)=-4\)

   គណនាអាំងតេក្រាលទូទៅ៖

   \[\begin{aligned} I(x) &= \int \left(\frac{3}{x+2} - \frac{4}{(x-1)^2}\right)dx \\ &= 3\ln|x+2| + \frac{4}{x-1} + C \end{aligned}\]

   ដោយ \(I(0) = -4 \implies 3\ln 2 - 4 + C = -4 \implies C = -3\ln 2\)

   ដូចនេះ៖ \(\boxed{I(x) = 3\ln\left|\frac{x+2}{2}\right| + \frac{4}{x-1}}\)

ដំណោះស្រាយ

1. កំណត់តម្លៃ \(A, B\) និង \(C\)

   តម្រូវភាគបែងរួម គេបាន៖

   \[\begin{aligned} 5x^2 - 14x + 13 &= A(x-3)^2 \\ &\quad{} + B(x+1)(x-3) \\ &\quad{} + C(x+1) \end{aligned}\]

   យក \(x = -1\)៖ \(5(-1)^2 - 14(-1) + 13 = 16A \implies A = 2\)

   យក \(x = 3\)៖ \(5(3)^2 - 14(3) + 13 = 4C \implies C = 4\)

   យក \(x = 0\)៖ \(13 = 9A - 3B + C \implies 3B = 9 \implies B = 3\)

   ដូចនេះ៖ \(\boxed{A = 2,\; B = 3,\; C = 4}\)

   គេបាន៖ \(\boxed{f(x) = \frac{2}{x+1} + \frac{3}{x-3} + \frac{4}{(x-3)^2}}\)

2. គណនាអាំងតេក្រាលមិនកំណត់ \(\displaystyle\int f(x) dx\)

   ធ្វើអាំងតេក្រាល៖

   \[\begin{aligned} I &= \int \left(\frac{2}{x+1} + \frac{3}{x-3} + \frac{4}{(x-3)^2}\right)dx \\ &= 2\ln|x+1| + 3\ln|x-3| - \frac{4}{x-3} + C \end{aligned}\]

   ដូចនេះ៖

   \(\boxed{I = 2\ln|x+1| + 3\ln|x-3| - \frac{4}{x-3} + C}\)

ដំណោះស្រាយ

1. រកតម្លៃ \(A, B\) និង \(C\)

   ដាក់ភាគបែងជាផលគុណកត្តា៖ \(x - 2x^2 + x^3 = x(1-x)^2\)

   តម្រូវភាគបែងរួម៖
   

   \[\begin{aligned} 4x^2 - 7x + 5 &= A(1-x)^2 + Bx(1-x) \\ &\quad{} + Cx \end{aligned}\]

   ជំនួស \(x = 0\)៖ \(5 = A(1)^2 \implies A = 5\)

   ជំនួស \(x = 1\)៖ \(4(1)^2 - 7(1) + 5 = C(1) \implies C = 2\)

   ផ្ទឹមមេគុណនៃ \(x^2\)៖ \(4 = A - B \implies B = A - 4 = 1\)

   ដូចនេះ៖ \(\boxed{A = 5,\; B = 1,\; C = 2}\)

   គេបាន៖ \(\boxed{f(x) = \frac{5}{x} + \frac{1}{1-x} + \frac{2}{(1-x)^2}}\)

2. គណនា \(k(x)=\displaystyle\int f(x) dx\) បើក្រាបតាង \(y=k(x)\) កាត់តាមចំណុច \(M(2,2)\)

   គណនាអាំងតេក្រាលទូទៅ៖

   \[\begin{aligned} k(x) &= \int \left(\frac{5}{x} + \frac{1}{1-x} + \frac{2}{(1-x)^2}\right)dx \\ &= 5\ln|x| - \ln|1-x| + \frac{2}{1-x} + C \end{aligned}\]

   ដោយ \(k(2) = 2 \implies 5\ln 2 - \ln 1 + \frac{2}{1-2} + C = 2 \implies 5\ln 2 - 2 + C = 2 \implies C = 4 - 5\ln 2\)

   ជំនួសតម្លៃ \(C\) គេបាន៖

   ដូចនេះ៖ \(\boxed{k(x) = 5\ln\left|\frac{x}{2}\right| - \ln|1-x| + \frac{2}{1-x} + 4}\)

ដំណោះស្រាយ

1. គណនាអាំងតេក្រាល \(I(x)=\displaystyle\int \left(\dfrac{e^{x}}{1-e^{x}}\right)dx\)

   តាង \(u = 1 - e^x \implies du = -e^x dx \implies e^x dx = -du\)

   \(I(x) = \displaystyle\int \frac{-du}{u} = -\ln|u| + C = -\ln|1-e^x| + C\)

   ដូចនេះ៖ \(\boxed{I(x) = -\ln|1-e^x| + C}\)

2. រកចំនួនពិត \(A\) និង \(B\)

   បំប្លែងភាគយក៖

   \[\frac{e^x}{1-e^x} = \frac{-(1-e^x) + 1}{1-e^x} = -1 + \frac{1}{1-e^x}\]

   ផ្ទឹមនឹងកន្សោម \(A + \dfrac{B}{1-e^x}\) គេទាញបាន៖

   ដូចនេះ៖ \(\boxed{A = -1,\quad{} B = 1}\)

3. ទាញរកអាំងតេក្រាល \(J(x)=\displaystyle\int \left(\dfrac{1}{1-e^{x}}\right) dx\)

   តាមលទ្ធផលសំណួរ (b) គេមាន \(\dfrac{1}{1-e^x} = \dfrac{e^x}{1-e^x} + 1\)

   \(J(x) = \displaystyle\int \left(\frac{e^x}{1-e^x} + 1\right) dx = I(x) + x + C'\)

   ដូចនេះ៖ \(\boxed{J(x) = x - \ln|1-e^x| + C}\)

ដំណោះស្រាយ

1. \(\displaystyle I = \int \frac{x^3-2x}{(x^2+1)^2}\,dx\)

   បំបែកភាគយក៖ \(x^3 - 2x = x(x^2+1) - 3x \implies \dfrac{x^3-2x}{(x^2+1)^2} = \dfrac{x}{x^2+1} - \dfrac{3x}{(x^2+1)^2}\)

   ដោយ \(d(x^2+1) = 2x\,dx\) គេបាន៖

   \[I = \frac{1}{2}\int \frac{d(x^2+1)}{x^2+1} - \frac{3}{2}\int (x^2+1)^{-2} d(x^2+1)\]

   ដូចនេះ៖ \(\boxed{I = \frac{1}{2}\ln(x^2+1) + \frac{3}{2(x^2+1)} + C}\)

2. \(\displaystyle J = \int \frac{x^7}{(1+x^4)^2}\,dx\)

   តាង \(u = 1+x^4 \implies x^4 = u-1\) និង \(du = 4x^3 dx \implies x^3 dx = \frac{1}{4}du\)

   \(J = \displaystyle\frac{1}{4}\int \frac{u-1}{u^2}du = \frac{1}{4}\int \left(\frac{1}{u} - u^{-2}\right)du\)

   ដូចនេះ៖ \(\boxed{J = \frac{1}{4}\ln(1+x^4) + \frac{1}{4(1+x^4)} + C}\)

3. \(\displaystyle K = \int \frac{x^2}{x^2-4x+3}\,dx\)

   ចែកពហុធា និងបំបែកជាប្រភាគងាយ៖

   \[\begin{aligned} \frac{x^2}{x^2-4x+3} &= 1 + \frac{4x-3}{(x-1)(x-3)} \\ &= 1 - \frac{1/2}{x-1} + \frac{9/2}{x-3} \end{aligned}\]

   ដូចនេះ៖ \(\boxed{K = x - \frac{1}{2}\ln|x-1| + \frac{9}{2}\ln|x-3| + C}\)