Вычислить интегралы ∫-8х²-4х-2 /-5хdx= ∫sin(-2x+5)dx= ∫x²lnxdx=. Created by bayguzinat. algebra-ru

-5хdx= ∫sin(-2x+5)dx= ∫x²lnxdx=...

bayguzinat @bayguzinat

July 2022 1 11 Report

Вычислить интегралы
∫-8х²-4х-2 /-5хdx=
∫sin(-2x+5)dx=
∫x²lnxdx=

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NNNLLL54

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$1)\; \; \int \frac{-8x^2-4x-2}{-5x} dx=\int (\frac{8}{5}x+\frac{4}{5}+\frac{2}{5}\cdot \frac{1}{x})dx=\\\\=\frac{8}{5}\cdot \frac{x^2}{2}+\frac{4}{5}x+\frac{2}{5}\cdot lnx+C\\\\2)\; \; \int sin(-2x+5)dx=[\, t=-2x+5,\; dt=-2dx\, ]=\\\\=-\frac{1}{2}\int sint\, dt=-\frac{1}{2}\cdot (-cost)+C=\frac{1}{2}cos(-2x+5)+C\\\\3)\; \; \int x^2lnx\, dx=[\, u=lnx,\; du= \frac{dx}{x} ,\; dv=x^2\, dx,\; v=\frac{x^3}{3}\, ]=$

$=uv-\int v\, du= \frac{x^3}{3} lnx-\int \frac{x^2\, dx}{3} = \frac{x^3}{3}lnx- \frac{x^3}{9}+C$

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