Question

интеграл неопределенный

Answer 1

$\displaystyle\int ln(x+\sqrt{1+x^2})dx=xln(x+\sqrt{1+x^2})-\int\frac{xdx}{\sqrt{1+x^2}}=\\=xln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C\\\\\\u=ln(x+\sqrt{1+x^2})=\ \textgreater \ du=\frac{1}{x+\sqrt{1+x^2}}*(1+\frac{x}{\sqrt{1+x^2}})dx=\\=\frac{dx}{\sqrt{1+x^2}}\\dv=dx=\ \textgreater \ v=x\\\\\int\frac{xdx}{\sqrt{1+x^2}}=\frac{1}{2}\int\frac{d(1+x^2)}{\sqrt{1+x^2}}=\sqrt{1+x^2}+C$