Высшая математика. Несобственные интегралы. Помогите, пожалуйста. Напишите решен...

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

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NNNLLL54

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

$2)\; \; \int \limits _1^{+\infty }\, x\, cosx\, dx=\lim\limits _{A \to +\infty}\int\limits^{A}_1\, x\, cosx\, dx=\\\\\\=\Big [\; \int x\cdot cosx\, dx=[\; u=x\; ,\; du=dx\; ,\; dv=cosx\, dx\; ,\; v=sinx\; ]=\\\\=uv-\int v\, du=x\cdot sinx-\int sinx\, dx=x\cdot sinx+cosx+C\; \Big ]=\\\\\\=\lim\limits _{A \to +\infty}\Big (x\cdot sinx+cosx\Big)\Big|_1^{A}=\lim\limits _{A \to +\infty}\Big(A\cdot sinA+cosA-sin1+cos1\Big)=\infty \\\\rasxoditsya$

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

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

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

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