Question

Высшая математика. Неопределенный интеграл

Answer 1

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

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