Question

.Помогите решить интегралы...

Answer 1

$intlimits_1^{e}, frac{dx}{xcdot ln^3x}=intlimits^{e}_1, (lnx)^{-3}cdot frac{dx}{x}=intlimits^{e}_1, (lnx)^{-3}cdot d(lnx)=\\=limlimits _{varepsilon to +0}intlimits^{e}_{1+varepsilon }, (lnx)^{-3}cdot d(lnx)=limlimits _{varepsilon to +0}frac{(lnx)^{-2}}{-2}Big |_{1+varepsilon }^{e}=\\=limlimits _{varepsilon to +0}Big (-frac{1}{2ln^2x}Big |_{1+varepsilon }^{e}Big )=-frac{1}{2}limlimits _{varepsilon to +0}Big (frac{1}{ln^2e}-frac{1}{ln^2(1+varepsilon )}Big )=$

$=-frac{1}{2}cdot (1-infty )=+infty quad [; lne=1; ,; ln1=0; ]\\\2); ; int underbrace {x}_{u}cdot underbrace {cosx, dx}_{dv}=[; du=dx; ,; v=sinx; ]=uv-int v, du=\\=xcdot sinx-int sinx, dx=xcdot sinx+cosx+C; ;\\intlimits_{frac{pi}{2} }^{+infty }, xcdot cosx, dx=limlimits _{B to +infty}int limits _{frac{pi }{2}}^{B}xcdot cosx, dx=limlimits _{B to +infty}(xcdot sinx+cosx)Big |_{frac{pi}{2}}^B=\\=limlimits _{B to +infty}(Bcdot sinB+cosB-frac{pi }{2}-0)=+infty$