Question

Интегралы .Номера 3.20, 3.21, 3,22, 3,23,3.4

Answer 1

$3.20) intlimits { frac{lnx}{ sqrt{x} } } , dx =|u=lnx||dv= frac{1}{ sqrt{x} } dx||du= frac{dx}{x}||v=2 sqrt{x} |=2 sqrt{x} *lnx \ - 2intlimits{ x^{- frac{1}{2} } } , dx=2 sqrt{x} *lnx-4 sqrt{x} +C \ 3.21) intlimits {(2x-3)sinfrac{x}{2} } , dx =|u=2x-3||dv=sin frac{x}{2}dx||du=2| \ |v=-2cos frac{x}{2} |=(2x-3)*(-2cos frac{x}{2})+4 intlimits {cos frac{x}{2} } , dx = \ =(2x-3)*(-2cos frac{x}{2})+8sin frac{x}{2}+C$

Answer 2

3.20.$int {frac{ln(x)}{sqrt{x}}},dx=left[begin{array}{cc}u=ln(x)&dv=frac{dx}{sqrt{x}}\du=frac{dx}{x}&v=2 sqrt{x}end{array}right]=2sqrt{x}*ln(x)-int {frac{2}{sqrt{x}}},dx=$$2sqrt{x}*ln(x)-4sqrt{x}+C$
3.21.$int {(2x-3)sinfrac{x}{2},dx=left[begin{array}{cc}u=2x-3&dv=sinfrac{x}{2}dx\du=2dx&v=-2cosfrac{x}{2}end{array}right]=$$2(3-2x)cosfrac{x}{2}+4intcosfrac{x}{2}, dx=2(3-2x)cosfrac{x}{2}+8sinfrac{x}{2}+C=$$2(3-2xcosfrac{x}{2}+4sinfrac{x}{2})+C$
3.22.$int{ln^2x},dx=left[begin{array}{cc}u=ln^2x&dv=dx\du=frac{2ln(x)dx}{x}&v=xend{array}right]=x*ln^2x-2int{ln(x)},dx=$$left[begin{array}{cc}u=ln(x)&dv=dx\du=frac{dx}{x}&v=xend{array}right]=x*ln^2x-2(x*ln(x)-int dx)=x*ln^2x-2(x*ln(x)-x)+C=x(ln^2x-2ln(x)+2)+C$
3.23.$int{frac{ln(x)}{(x+1)^2}},dx=left[begin{array}{cc}u=ln(x)&dv=frac{dx}{(x+1)^2} \du=frac{dx}{x}&v=-frac{1}{x+1}end{array}right]=-frac{ln(x)}{x+1}+int{frac{dx}{x(x+1)}=-frac{ln(x)}{x+1}+int{frac{dx}{x}-int{frac{dx}{x+1}$$=-frac{ln(x)}{x+1}+ln(x)-ln(x+1)+C$
3.24.$int{cos(ln(x))}dx=left[begin{array}{cc}u=cos(ln(x))&dv=dx\du=-frac{sin(ln(x))}{x}dx&v=x}end{array}right]=$$x*cos(ln(x))+int{sin(ln(x))}dx=left[begin{array}{cc}u=sin(ln(x))&dv=dx\du=frac{cos(ln(x))}{x}dx&v=x}end{array}right]=$$=x*cos(ln(x))+x*sin(ln(x))-int cos(ln(x))dx\int cos(ln(x))dx=frac{x(sin(ln(x))+cos(ln(x)))}{2}+C$