помогите пожалуйста . Created by Allena2000. matematika-ru

помогите пожалуйста ...

Allena2000 @Allena2000

July 2022 1 1 Report

помогите пожалуйста

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nafanya2014

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$>u=1-4xdu=(1-4x)`dxdu=-4dxd(1-4x)=-4dx<img src=$

Аналогично

$\int \sqrt[3]{x}dx \,=\frac{x^{\frac{4}{3} }}{\frac{4}{3} } +C\\ \\ \int \sqrt[3]{u}du ={\frac{3u^{\frac{4}{3} }}{4} } \+C$

u=x²+1

du=2xdx

d(x^2+1)=2xdx

$\int x\sqrt[3]{x^2+1}dx =\frac{1}{2}\int \sqrt[3]{x^2+1}\cdot(2xdx)=\frac{1}{2}\int \sqrt[3]{x^2+1}\cdot d(x^2+1)= {\frac{3(x^2+1)^{\frac{4}{3} }}{4} } +C$

$\int cosxdx=sinx+C\\ \\ \int cosu du=sinu+C$

u=(x/6)-1

du=dx/6

$\int cos(\frac{x}{6}-1) dx= 6\int cos(\frac{x}{6}-1) \frac{dx}{6} =6sin(\frac{x}{6}-1) +C$

$\int e^{x}dx=e^{x}+C\\ \\ \int e^{u}du=e^{u}+C$

u=√x

du=dx/(2√x)

$\int e^{\sqrt{x} }dx=2 \int \frac{e^{\sqrt{x}dx }}{2\sqrt{x} }=2e^{\sqrt{x} }+C$

$\int {\frac{dx}{x^2-a^2} } =\frac{1}{2a}ln|\frac{x-a}{x+a}|+C$

u=x³

du=3x²dx

$\int {\frac{x^2dx}{x^6-4} } =\frac{1}{3}\int {\frac{3x^2dx}{(x^3)^2-2^2} }=\frac{1}{3}\int {\frac{d(x^3)}{(x^3)^2-2^2} }=\frac{1}{3} \cdot \frac{1}{2\cdot 2}ln|\frac{x^2-2}{x^3+2}|+C = \frac{1}{12}ln|\frac{x^2-2}{x^3+2}|+C$

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