Question

Помогите с решением интегралов. Заранее буду благодарен. Сделайте с подробным решением.

Answer 1

$1. int{x^4}, dx=frac{x^5}{5}+C$
$2. int{x^{-3}}, dx=frac{x^{-2}}{-2}+C=-frac{1}{2x^2}+C$
$3. int{5}, dx=5int{dx}=5x+C$
$4. int{frac{1}{2}t^2}, dt=frac{1}{2}int{t^2}, dt=frac{1}{2}*frac{t^3}{3}+C=frac{t^3}{6}+C$
$5. int{(3x-x^2)}, dx=int{3x}, dx-int{x^2}, dx=frac{3x^2}{2}-frac{x^3}{3}+C$
$6. int{4x^3+4x-3}, dx=4int{x^3}, dx+4int{x}, dx-3int{dx}=$
$=frac{4x^4}{4}+frac{4x^2}{2}-3x+C=x^4+2x^2-3x+C$
$7. int{x^2(1+2x)}, dx=int{(x^2+2x^3)}, dx=int{x^2}, dx+2int{x^3}, dx=$
$=frac{x^3}{3}+frac{2x^4}{4}+C=frac{x^3}{3}+frac{x^4}{2}+C$
$8. int{4(2x-1)^2}, dx=int{4(4x^2-4x+1)}, dx=int{(16x^2-16x+4)}, dx=$
$=16int{x^2}, dx-16int{x}, dx+4int{}, dx=frac{16x^3}{3}-frac{16x^2}{2}+4x+C=$
$=frac{16x^3}{3}-8x^2+4x+C$
$9. int{x(1-x)^2}, dx=int{x(1-2x+x^2)}, dx=int{(x-2x^2+x^3)}, dx=$
$=int{x}, dx-2int{x^2}, dx+int{x^3}, dx=frac{x^2}{2}-frac{2x^3}{3}+frac{x^4}{4}+C$
$10. int{sqrt[7]{x^4}}, dx=int{x^{frac{4}{7}}}, dx=frac{x^{frac{4}{7}+1}}{frac{4}{7}+1}+C=frac{x^{frac{11}{7}}}{frac{11}{7}}+C=frac{7x^{frac{11}{7}}}{11}+C=$
$=frac{7}{11}x^{frac{11}{7}}+C=frac{7}{11}sqrt[7]{x^{11}}+C=frac{7}{11}xsqrt[7]{x^4}+C$
$11. int{frac{1}{x^{-3}}}, dx=int{frac{1}{frac{1}{x^3}}}, dx=int{x^3}, dx=frac{x^4}{4}+C$
$12.int{frac{dx}{sqrt{x}}}=int{frac{1}{sqrt{x}}}, dx=2sqrt{x}+C$
$13.int{frac{3dx}{4sqrt[3]{x}}}=frac{3}{4}int{frac{1}{sqrt[3]{x}}}, dx=frac{3}{4}int{frac{1}{x^{frac{1}{3}}}}, dx=frac{3}{4}int{x^{-frac{1}{3}}}, dx=$
$=frac{3}{4}*frac{x^{1-frac{1}{3}}}{1-frac{1}{3}}+C=frac{3}{4}*frac{x^frac{2}{3}}{frac{2}{3}}+C=frac{3}{4}*frac{3sqrt[3]{x^2}}{2}+C=frac{9sqrt[3]{x^2}}{8}+C$
$14. int{frac{xdx}{3xsqrt{x}}}=int{frac{1}{3sqrt{x}}}, dx=frac{1}{3}int{frac{1}{sqrt{x}}}, dx=frac{1}{3}*2sqrt{x}+C=frac{2sqrt{x}}{3}+C$
$15. int{(frac{3}{x^2}-frac{5}{3sqrt{x}}+2sqrt[3]{x^2})}, dx=3int{frac{1}{x^2}}, dx-frac{5}{3}int{frac{1}{sqrt{x}}}, dx+2int{x^{frac{2}{3}}}, dx=$
$=3*(-frac{1}{x})-frac{5}{3}*2sqrt{x}+2*frac{3x^{frac{5}{3}}}{5}+C=$
$=-frac{3}{x}-frac{10sqrt{x}}{3}+frac{6sqrt[3]{x^5}}{5}+C=frac{6sqrt[3]{x^5}}{5}-frac{10sqrt{x}}{3}-frac{3}{x}+C$
$16. int{frac{x-sqrt[3]{x^2}}{sqrt{x}}}, dx=$
--------------------------------
Замена:
$u=sqrt{x}$
$du=frac{1}{2sqrt{x}}, dx$
-------------------------------------
$=2int{(u^2-(u^2)^{frac{2}{3}}))}, du=2int{u^2}, du-2int{u^{frac{4}{3}}}, du=frac{2u^3}{3}-2*frac{x^{frac{7}{3}}}{frac{7}{3}}+C=$
$=frac{2u^3}{3}-frac{6u^{frac{7}{3}}}{7}+C=frac{2sqrt{x^3}}{3}-frac{6sqrt[6]{x^7}}{7}+C$

$17. int{frac{1}{2}cosx}, dx=frac{1}{2}int{cosx}, dx=frac{sinx}{2}+C$

$18. int{frac{3}{2sin^2x}}, dx=frac{3}{2}int{frac{1}{sin^2x}}, dx=-frac{3ctgx}{2}+C$

$19. int{(1+cosx)}, dx=int{}, dx+int{cosx}, dx=x+sinx+C$

$20. int{(4x^2+2cosx)}, dx=4int{x^2}, dx+2int{cosx}, dx=frac{4x^3}{3}+2sinx+C$

$21. int{(frac{2}{cos^2x}-frac{3}{sin^2x})}, dx=2int{frac{1}{cos^2x}}, dx-3int{frac{1}{sin^2x}}, dx=$
$=2tgx+3ctgx+C$

$22. int{3e^u}, du=3int{e^u}, du=3e^u+C$

$23. int{(x-5e^x)}, dx=int{x}, dx-5int{e^x}, dx=frac{x^2}{2}-5e^x+C$

$24. int{(2e^t-3cos,t)}, dt=2int{e^t}, dt-3int{cos, t}, dt=2e^t-3sin,t+C$

$25. int{(frac{2}{x}-x)}, dx=2int{frac{1}{x}}, dx-int{x}, dx=2ln|x|-frac{x^2}{2}+C$

$26. int{frac{6dx}{1+x^2}}=6int{frac{1}{1+x^2}}, dx=6arctgx+C$


$27. int{frac{2cos^2v+1}{cos^2v}}, dv=int{((2cos^2v+1)*sec^2v)}, dv=$
$=int{(sec^2v+2)}, dv=int{sec^2v}, dv+2int{}, dv=tg, v+2v+C$