Question

Прошу решите интегралы!

Answer 1

Ответ:

2. ∫xdx=$\frac{x^{2} }{2}$+C

3. ∫$x^{6}$dx=$\frac{x^{7} }{7}$+C

4. ∫3dx=3x+C

5.∫5xdx=$\frac{5x^{2} }{2}$+C

6. ∫(4-x)dx=4∫dx-∫xdx=4x-$\frac{x^{2} }{2}$+C

7. ∫(5x-$x^{2}$)dx=5∫xdx-∫$x^{2}$dx=$\frac{5x^{2} }{2}-\frac{x^{3} }{3} +C$

8. ∫3(x-3)dx=∫(3x-9)dx=3∫xdx-9∫dx=$\frac{3x^{2} }{2} -9x+C$

9.∫$(4x^{3} +8x-2)dx=$ 4∫$x^{3}$dx+8∫xdx-2∫dx=$\frac{4x^{4} }{4} + \frac{8x^{2} }{2} -2x=x^{4} +4x^{2} -2x+C$

10. ∫$x^{2} (1+4x)dx=$ ∫$(x^{2} +4x^{3} )dx=$∫$x^{2} dx$+4∫$x^{3} dx$=$\frac{x^{3} }{3} +\frac{4x^{4} }{4} =\frac{x^{3} }{3} +x^{4} +C$

11. ∫$4(3x-2)^2dx=$∫4 × $(9x^{2} -12x+4)$dx=∫(36$x^{2}$-48x+16)dx=36∫$x^{2}$dx-48∫xdx+16∫dx=$\frac{36x^{3} }{3} -\frac{48x^{2} }{2} +16x$=$12x^{3} -24x^{2} +16x+C$

12. ∫$x(5-x)^2$dx=∫$(x^{3} -10x^{2} +25x)dx$=∫$x^{3}$dx-10∫$x^{2}$dx+25∫xdx=$\frac{x^{4} }{4} -\frac{10x^{3} }{3} +\frac{25x^{2} }{2} +C$

13.∫$2\sqrt{x} dx=\frac{4x\sqrt{x} }{3} +C$

14. ∫  $\frac{dx}{x^{2} }= \frac{1}{\frac{x^{3} }{3} } =\frac{3}{x^{3} } +C$

15. ∫$\frac{dx}{\sqrt[3]{x^{2} } }$= ∫$x^-\frac{2}{3}$dx=$\frac{x^{\frac{1}{3} } }{\frac{1}{3} } =3\sqrt[3]{x} +C$

16.∫$(\frac{2}{x^{2} } -\frac{4}{\sqrt{x} } +3\sqrt[3]{x^{2} })$=∫$\frac{2}{x^{2} } -\frac{4}{x^{\frac{1}{2} } }+3x^{\frac{2}{3} } dx$=∫$\frac{2}{x^{2} }$dx-∫$\frac{4}{x^{\frac{1}{2} } }$dx+∫3$x^{\frac{2}{3} }$dx= -$\frac{2}{x} - 8\sqrt{x} +\frac{9x\sqrt[3]{x^{2} } }{5} +C$

17.∫2cosxdx=2∫cosxdx=2sinx+C

18.∫$\frac{3dx}{sin^{2}x }$=3×∫$\frac{1}{sin^{2} x}$dx=-3ctgx+C

19. ∫(1+cosx)dx=∫dx+∫cosxdx=x+sinx+C

20.∫3$x^{2}$-2cosx)dx=3∫$x^{2}$dx-2∫cosxdx=$\frac{3x^{3} }{3} -2sinx+C$

21. ∫$(\frac{2}{x} -x)$dx=∫$\frac{2}{x}$dx-∫xdx=2㏑(|x|)-$\frac{x^{2} }{2}$+C