Question

Синтегралами. 1)интеграл x^2dx/корень7-x^2 2)интеграл dx/x^2корень1+x^2

Answer 1

Скобки...
$$$ \large \int\frac{x^2dx}{\sqrt{7-x^2}}=-\int\frac{7-x^2-7}{\sqrt{7-x^2}}=-\int\sqrt{7-x^2}dx+7\int\frac{dx}{\sqrt{7-x^2}}=-\\-x\sqrt{7-x^2}-\int\frac{x^2dx}{\sqrt{7-x^2}}+7\int\frac{dx}{\sqrt{7-x^2}}\\\\ \int\frac{x^2dx}{\sqrt{7-x^2}}=-\frac{x}{2}\sqrt{7-x^2}+\frac{7}{2}arcsin\frac{x}{\sqrt{7}}+C\\\\\\\int\sqrt{7-x^2}dx=x\sqrt{7-x}+\int\frac{x^2dx}{\sqrt{7-x^2}}\\u=\sqrt{7-x^2};du=-\frac{xdx}{\sqrt{7-x^2}}\\dv=dx;v=x$

$$$ \large \int\frac{dx}{x^2\sqrt{1+x^2}}=-\int dt=-t+C=-\frac{\sqrt{x^2+1}}{x}+C\\\\\\m=-2;n=2;p=-\frac{1}{2};a=1;b=1\\\\1+\frac{1}{x^2}=t^2;x^2=\frac{1}{t^2-1}\\\sqrt{1+x^2}=\sqrt{1+\frac{1}{t^2-1}}=\frac{t}{\sqrt{t^2-1}}\\\frac{dx}{x^2};-\frac{dx}{x^3}=tdt;\frac{dx}{x^2}=-txdt=-\frac{tdt}{\sqrt{t^2-1}}$