Ребят помогите найти интегралы ∫(7x⁶-sinx+3)dx ∫(5-∛x²)/x ∫(3x-5)⁶dx ∫(sinx dx)/3-cosx ∫(xdx)/(x²+3)². Created by mkineshma. algebra-ru

(x²+3)²...

mkineshma @mkineshma

March 2022 1 6 Report

Ребят помогите найти интегралы
∫(7x⁶-sinx+3)dx
∫(5-∛x²)/x
∫(3x-5)⁶dx
∫(sinx dx)/3-cosx
∫(xdx)/(x²+3)²

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NNNLLL54

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$1)\; \int (7x^6-sinx+3)dx=7\cdot \frac{x^7}{7}+cosx+3x+C\\\\2)\; \int (5-\sqrt[3]{x^2})dx=5x-\frac{3x^{\frac{5}{3}}}{5}+C\\\\3)\; \int (3x-5)^6dx=\frac{1}{3}\cdot \frac{(3x-5)^7}{7}+C$

$4)\; \int \frac{sinx\, dx}{3-cosx} =[\, t=3-cosx,\; dt=sinx\, dx\, ]=\int \frac{dt}{t}=\\\\=ln|t|+C=ln|3-cosx|+C\\\\5)\; \int \frac{x\, dx}{(x^2+3)^2} =[\, t=x^2+3,\; dt=2t\, dt\, ]=\frac{1}{2}\int \frac{dt}{t^2}=\\\\=\frac{1}{2}\cdot (-\frac{1}{t})+C=-\frac{1}{2}\cdot \frac{1}{x^2+3}+C$

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