Задача №1423

\[ \int\sqrt{a^2+x^2}dx =\left[\begin{aligned}& u=\sqrt{a^2+x^2};\;du=\frac{xdx}{\sqrt{a^2+x^2}}.\\& dv=dx;\;v=x.\end{aligned}\right] =x\cdot\sqrt{a^2+x^2}-\int{x\cdot\frac{xdx}{\sqrt{a^2+x^2}}}=\\ =x\cdot\sqrt{a^2+x^2}-\int\frac{x^2}{\sqrt{a^2+x^2}}dx =x\cdot\sqrt{a^2+x^2}-\int\frac{x^2+a^2-a^2}{\sqrt{a^2+x^2}}dx=\\ =x\cdot\sqrt{a^2+x^2}-\int\left(\sqrt{a^2+x^2}-\frac{a^2}{\sqrt{a^2+x^2}}\right)dx =x\cdot\sqrt{a^2+x^2}-\int\sqrt{a^2+x^2}dx+a^2\int\frac{dx}{\sqrt{a^2+x^2}}=\\ =x\cdot\sqrt{a^2+x^2}-\int\sqrt{a^2+x^2}dx+a^2\ln\left(x+\sqrt{x^2+a^2}\right). \]

\[ \int\sqrt{a^2+x^2}dx =x\cdot\sqrt{a^2+x^2}-\int\sqrt{a^2+x^2}dx+a^2\ln\left(x+\sqrt{x^2+a^2}\right);\\ \int\sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\ln\left(x+\sqrt{x^2+a^2}\right)+C =\frac{x\sqrt{a^2+x^2}}{2}+\frac{a^2\ln\left(x+\sqrt{x^2+a^2}\right)}{2}+C \]