Question

Высшая математика. Неопределенный интеграл

Answer 1

$\displaystyle\int\frac{lnx}{\sqrt{(1+x^2)^3}}=\frac{xlnx}{\sqrt{1+x^2}}-\int\frac{dx}{\sqrt{1+x^2}}=\frac{xlnx}{\sqrt{1+x^2}}-ln|x+\sqrt{1+x^2}|+C\\\\u_1=lnx;du_1=\frac{dx}{x}\\dv_1=\frac{dx}{\sqrt{(1+x^2)^3}};v_1=\frac{x}{\sqrt{1+x^2}}\\\int\frac{dx}{\sqrt{(1+x^2)^3}}=\int\frac{(1+x^2-x^2)dx}{\sqrt{(1+x^2)^3}}=\int\frac{dx}{\sqrt{1+x^2}}-\int\frac{x^2}{\sqrt{(1+x^2)^3}}dx=\\=\int\frac{dx}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}-\int\frac{dx}{\sqrt{1+x^2}}=\frac{x}{\sqrt{1+x^2}}$

$\displaystyle\int\frac{x^2}{\sqrt{(1+x^2)^3}}dx\\u_2=x;du_2=dx\\dv_2=\frac{xdx}{\sqrt{(1+x^2)^3}};v_2=-\frac{1}{\sqrt{1+x^2}}$