Question

7 Интегралов помогите срочно

Answer 1

(12)
$\int\limits {\frac{[4-ln(x)]^2}{x}} \, dx = \int\limits {[4-ln(x)]^2*\frac{1}{x}} \, dx =\\\\ = \int\limits {[4-ln(x)]^2} \, d[ln(x)] = -\int\limits {[4-ln(x)]^2} \, d[-ln(x)] =\\\\ = -\int\limits {[4-ln(x)]^2} \, d[4-ln(x)] =-\frac{[4-ln(x)]^3}{3}+C$

(13)
$\int\limits {cos(x-2)} \, dx = \int\limits {cos(x-2)} \, d(x-2) =\\\\ =sin(x-2)+C$

(14)
$\int\limits {xcos(2x)} \, dx =\frac{1}{2} \int\limits {x*2cos(2x)} \, dx =\\\\ =\frac{1}{2} \int\limits {x} \, d(sin(2x)) =\frac{1}{2}*[x*sin(2x)- \int\limits {sin(2x)} \, dx ]=\\\\ =\frac{1}{2}*[x*sin(2x)- \frac{1}{2}*\int\limits {sin(2x)} \, d(2x) ]=\\\\ =\frac{1}{2}*[x*sin(2x)+ \frac{1}{2}*cos(2x)]+C=\\\\ =\frac{x*sin(2x)}{2}+\frac{cos(2x)}{4}+C$

(15)
$\int\limits {ln^2(x)} \, dx =x*ln^2(x)- \int\limits {x*} \, d[ln^2(x)] =\\\\ =x*ln^2(x)- \int\limits {x*2ln(x)*\frac{1}{x}} \, dx =\\\\ =x*ln^2(x)- 2*\int\limits {ln(x)} \, dx =\\\\ =x*ln^2(x)- 2*[x*ln(x)-\int\limits {x} \, d[ln(x)]] =\\\\ =x*ln^2(x)- 2*[x*ln(x)-\int\limits {x*\frac{1}{x}} \, dx] =\\\\ =x*ln^2(x)- 2x*ln(x)+2\int\limits {1} \, dx=\\\\ =x*ln^2(x)- 2x*ln(x)+2x+C=\\\\ =x*(ln^2(x)- 2ln(x)+2)+C$

(16)
$\int\limits {xe^{-2x}} \, dx = \int\limits {x} \, d(\frac{e^{-2x}}{-2}) =-\frac{1}{2}* \int\limits {x} \, d(e^{-2x}) =\\\\ =-\frac{1}{2}*[x*e^{-2x}- \int\limits {e^{-2x}} \, dx ]=\\\\ =-\frac{1}{2}*[x*e^{-2x}+\frac{1}{2} *\int\limits {e^{-2x}} \, d(-2x) ]=\\\\ =-\frac{1}{2}*[x*e^{-2x}+\frac{1}{2} *e^{-2x} ]+C=\\\\ =-\frac{e^{-2x}}{2}*[x+\frac{1}{2}]+C=\\\\ =-\frac{2x+1}{4}*e^{-2x}+C$

(17)
$\int\limits {(2x-3)*sin(\frac{x}{2})} \, dx = \int\limits {(2x-3)} \, d[-2cos(\frac{x}{2})] =\\\\ =-2cos(\frac{x}{2})*(2x-3)- \int\limits {[-2cos(\frac{x}{2})]} \, d(2x-3) =\\\\ =-2cos(\frac{x}{2})*(2x-3)+4\int\limits {cos(\frac{x}{2})} \, dx =\\\\ =-2cos(\frac{x}{2})*(2x-3)+8\int\limits {cos(\frac{x}{2})} \, d(\frac{x}{2}) =\\\\ =-2cos(\frac{x}{2})*(2x-3)+8sin(\frac{x}{2})+C=\\\\ =(6-4x)*cos(\frac{x}{2})+8sin(\frac{x}{2})+C$

(18)
$\int\limits {(x^2+1)e^x} \, dx = \int\limits {(x^2+1)} \, d(e^x) =\\\\ =e^x*(x^2+1)- \int\limits {e^x*(x^2+1)'} \, dx =\\\\ =e^x*(x^2+1)- \int\limits {2x*e^x} \, dx =\\\\ =e^x*(x^2+1)- \int\limits {2x} \, d(e^x) =\\\\ =e^x*(x^2+1)- 2x*e^x+ \int\limits {e^x} \, d(2x) =\\\\ =e^x*(x^2+1- 2x)+ 2\int\limits {e^x} \, dx =\\\\ =e^x*(x^2- 2x+1)+ 2e^x+C =\\\\ =e^x*(x^2- 2x+1+2)+ C =\\\\ =e^x*(x^2- 2x+3)+ C$