Помогите пожалуйста!!!!. Created by DolcheGabanna. algebra-ru

Помогите пожалуйста!!!!...

DolcheGabanna @DolcheGabanna

August 2022 1 5 Report

Помогите пожалуйста!!!!

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NNNLLL54

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$1)\ \ \int (10x^4+30-8x-24x^{-5})\, dx=2x^5+30x-4x^2+6x^{-4}+C\\\\2)\ \ \int (2x-5)(5+2x)\, dx=\int (4x^2-25)\, dx=\dfrac{4x^3}{3}-25x+C\\\\3)\ \ \int (4cosx-8sinx)\, dx=4sinx+8cosx+C\\\\4)\ \ \int \Big(\dfrac{2}{cos^2x}+2x\Big)\, dx=2tgx+x^2+C\\\\5)\ \ \int \Big(1-\dfrac{4}{sin^2x}\Big)\, dx=x+4ctgx+C$

$6)\ \ \int sin(\pi -x)\, dx=cos(\pi -x)+C=-cosx+C\\\\7)\ \ \int \Big(8\cdot 4^{x}+\dfrac{2}{9}\, e^{x}\Big)\, dx=8\cdot \dfrac{4^{x}}{ln4}+\dfrac{2}{9}\, e^{x}+C\\\\8)\ \ \int \Big(\dfrac{1}{5x}+\dfrac{1}{x^5}\Big)\, dx=\dfrac{1}{5}\, ln|x|-\dfrac{1}{4x^4}+C\\\\9)\ \ \int \Big(\dfrac{-4}{1+x^2}-1\Big)\, dx=-4\cdot arctgx-x+C\\\\10)\ \ \int \Big(\dfrac{9}{\sqrt{1-x^2}}+3x\Big)\, dx=9\cdot arcsinx+\dfrac{3x^2}{2}+C$

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