Question

8.2.22 помогите интеграл решить

Answer 1

$intlimits {(2x-1)e^{3x}} , dx = intlimits {(2x-1)} , d( frac{e^{3x}}{3} ) = frac{1}{3} intlimits {(2x-1)} , d(e^{3x}) =$

$= frac{1}{3}*[(2x-1)e^{3x}- intlimits {e^{3x}} , d(2x-1)] =$

$= frac{1}{3}*[(2x-1)e^{3x}- intlimits {e^{3x}} , d(2x)] = frac{1}{3}*[(2x-1)e^{3x}- 2 intlimits {e^{3x}} , dx] =$

$= frac{1}{3}*[(2x-1)e^{3x}- 2 intlimits {e^{3x}} , d (frac{3x}{3} )]=$

$= frac{1}{3}*[(2x-1)e^{3x}- frac{2}{3} intlimits {e^{3x}} , d(3x)] = frac{1}{3}*[(2x-1)e^{3x}- frac{2}{3} e^{3x}]+C =$

$= frac{e^{3x}}{3}*[2x-1-frac{2}{3}]+C = frac{e^{3x}}{3}*(2x-frac{5}{3})+C$