Вопрос в картинках...

Решите задачу:

$\LARGE \int_{0}^{1}{\sqrt{x}\over 1+x}{\mathrm{d}x}=\left [ u^2=x, {\mathrm{d}x}=2u{\mathrm{d}u} \right ]=2\int_{0}^{1}{u^2\over 1+u}{\mathrm{d}u} = 2\left ( \int_{0}^{1}{u^2+1-1\over 1+u^2}{\mathrm{d}u} \right )=2\left (\int_{0}^{1} {\mathrm{d}u}-\int_{0}^{1}{1\over 1+u^2}{\mathrm{d}u} \right )=2u|_{0}^{1}-2arctgu|_{0}^{1}=2-{\pi \over 2}$