Question

Высшая математика. Несобственные интегралы.
Помогите, пожалуйста. Напишите решение на бумаге.

Answer 1

$1); ; intlimits^{+infty }_2, dfrac{dx}{xcdot lnx}=limlimits _{A to +infty}intlimits^{A}_2Big(dfrac{1}{lnx}cdot dfrac{dx}{x}Big)=limlimits _{A to +infty}intlimits^{A}_2Big(dfrac{1}{lnx}cdot d(lnx)Big)=\\\=limlimits _{A to +infty}ln|lnx|, Big|_2^{A}=limlimits _{A to +infty}Big(underbrace {ln|lnA|}_{to ; +infty }-underbrace {ln|ln2|}_{const}Big)=+infty ; ; ,; ; rasxoditsya$

$2); ; int limits _1^{+infty }, x, cosx, dx=limlimits _{A to +infty}intlimits^{A}_1, x, cosx, dx=\\\=Big [; int xcdot cosx, dx=[; u=x; ,; du=dx; ,; dv=cosx, dx; ,; v=sinx; ]=\\=uv-int v, du=xcdot sinx-int sinx, dx=xcdot sinx+cosx+C; Big ]=\\\=limlimits _{A to +infty}Big (xcdot sinx+cosxBig)Big|_1^{A}=limlimits _{A to +infty}Big(Acdot sinA+cosA-sin1+cos1Big)=infty \\rasxoditsya$

$3); ; intlimits^2_1, dfrac{x, dx}{sqrt{x^2-1}}=frac{1}{2}cdot limlimits _{varepsilon to 0}intlimits^2_{1+varepsilon }, dfrac{2x, dx}{sqrt{x^2-1}}=frac{1}{2}cdot limlimits _{varepsilon to 0}Big(2sqrt{x^2-1}Big|_{1+varepsilon }^2Big)=\\\=frac{1}{2}cdot limlimits _{varepsilon to 0} Big(2sqrt{3}-2underbrace {sqrt{2varepsilon +varepsilon ^2}}_{to ; 0}}Big)=sqrt3; ; ; sxoditsya$

$4); ; intlimits^1_0, dfrac{dx}{x^2}}=limlimits _{varepsilon to +0}intlimits^1_{varepsilon }, dfrac{dx}{x^2}=limlimits _{varepsilon to +0}Big(-dfrac{1}{x}, Big|_{varepsilon }^1Big)=limlimits _{varepsilon to +0} Big(-dfrac{1}{1}+dfrac{1}{varepsilon }Big)=+infty ; ; ; rasxoditsya$

$5); ; intlimits^1_0, dfrac{dx}{sqrt{x}}}=limlimits _{varepsilon to +0}intlimits^1_{varepsilon }, dfrac{dx}{sqrt{x}}=limlimits _{varepsilon to +0}Big(2sqrt{x}, Big|_{varepsilon }^1Big)=limlimits _{varepsilon to +0} Big(2sqrt{1}-2sqrt{{varepsilon }}Big)=2; ; ; sxoditsya$