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

July 2022 1 8 Report

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

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o1l7eg17 $1. \int{x^4}\, dx=\frac{x^5}{5}+C$
$2. \int{x^{-3}}\, dx=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C$
$3. \int{5}\, dx=5\int{dx}=5x+C$
$4. \int{\frac{1}{2}t^2}\, dt=\frac{1}{2}\int{t^2}\, dt=\frac{1}{2}*\frac{t^3}{3}+C=\frac{t^3}{6}+C$
$5. \int{(3x-x^2)}\, dx=\int{3x}\, dx-\int{x^2}\, dx=\frac{3x^2}{2}-\frac{x^3}{3}+C$
$6. \int{4x^3+4x-3}\, dx=4\int{x^3}\, dx+4\int{x}\, dx-3\int{dx}=$
$=\frac{4x^4}{4}+\frac{4x^2}{2}-3x+C=x^4+2x^2-3x+C$
$7. \int{x^2(1+2x)}\, dx=\int{(x^2+2x^3)}\, dx=\int{x^2}\, dx+2\int{x^3}\, dx=$
$=\frac{x^3}{3}+\frac{2x^4}{4}+C=\frac{x^3}{3}+\frac{x^4}{2}+C$
$8. \int{4(2x-1)^2}\, dx=\int{4(4x^2-4x+1)}\, dx=\int{(16x^2-16x+4)}\, dx=$
$=16\int{x^2}\, dx-16\int{x}\, dx+4\int{}\, dx=\frac{16x^3}{3}-\frac{16x^2}{2}+4x+C=$
$=\frac{16x^3}{3}-8x^2+4x+C$
$9. \int{x(1-x)^2}\, dx=\int{x(1-2x+x^2)}\, dx=\int{(x-2x^2+x^3)}\, dx=$
$=\int{x}\, dx-2\int{x^2}\, dx+\int{x^3}\, dx=\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+C$
$10. \int{\sqrt[7]{x^4}}\, dx=\int{x^{\frac{4}{7}}}\, dx=\frac{x^{\frac{4}{7}+1}}{\frac{4}{7}+1}+C=\frac{x^{\frac{11}{7}}}{\frac{11}{7}}+C=\frac{7x^{\frac{11}{7}}}{11}+C=$
$=\frac{7}{11}x^{\frac{11}{7}}+C=\frac{7}{11}\sqrt[7]{x^{11}}+C=\frac{7}{11}x\sqrt[7]{x^4}+C$
$11. \int{\frac{1}{x^{-3}}}\, dx=\int{\frac{1}{\frac{1}{x^3}}}\, dx=\int{x^3}\, dx=\frac{x^4}{4}+C$
$12.\int{\frac{dx}{\sqrt{x}}}=\int{\frac{1}{\sqrt{x}}}\, dx=2\sqrt{x}+C$
$13.\int{\frac{3dx}{4\sqrt[3]{x}}}=\frac{3}{4}\int{\frac{1}{\sqrt[3]{x}}}\, dx=\frac{3}{4}\int{\frac{1}{x^{\frac{1}{3}}}}\, dx=\frac{3}{4}\int{x^{-\frac{1}{3}}}\, dx=$
$=\frac{3}{4}*\frac{x^{1-\frac{1}{3}}}{1-\frac{1}{3}}+C=\frac{3}{4}*\frac{x^\frac{2}{3}}{\frac{2}{3}}+C=\frac{3}{4}*\frac{3\sqrt[3]{x^2}}{2}+C=\frac{9\sqrt[3]{x^2}}{8}+C$
$14. \int{\frac{xdx}{3x\sqrt{x}}}=\int{\frac{1}{3\sqrt{x}}}\, dx=\frac{1}{3}\int{\frac{1}{\sqrt{x}}}\, dx=\frac{1}{3}*2\sqrt{x}+C=\frac{2\sqrt{x}}{3}+C$
$15. \int{(\frac{3}{x^2}-\frac{5}{3\sqrt{x}}+2\sqrt[3]{x^2})}\, dx=3\int{\frac{1}{x^2}}\, dx-\frac{5}{3}\int{\frac{1}{\sqrt{x}}}\, dx+2\int{x^{\frac{2}{3}}}\, dx=$
$=3*(-\frac{1}{x})-\frac{5}{3}*2\sqrt{x}+2*\frac{3x^{\frac{5}{3}}}{5}+C=$
$=-\frac{3}{x}-\frac{10\sqrt{x}}{3}+\frac{6\sqrt[3]{x^5}}{5}+C=\frac{6\sqrt[3]{x^5}}{5}-\frac{10\sqrt{x}}{3}-\frac{3}{x}+C$
$16. \int{\frac{x-\sqrt[3]{x^2}}{\sqrt{x}}}\, dx=$
--------------------------------
Замена:
$u=\sqrt{x}$
$du=\frac{1}{2\sqrt{x}}\, dx$
-------------------------------------
$=2\int{(u^2-(u^2)^{\frac{2}{3}}))}\, du=2\int{u^2}\, du-2\int{u^{\frac{4}{3}}}\, du=\frac{2u^3}{3}-2*\frac{x^{\frac{7}{3}}}{\frac{7}{3}}+C=$
$=\frac{2u^3}{3}-\frac{6u^{\frac{7}{3}}}{7}+C=\frac{2\sqrt{x^3}}{3}-\frac{6\sqrt[6]{x^7}}{7}+C$

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

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

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

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

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

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

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

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

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

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

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

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