Интегралот1 до0 x / (4+x^2)

Внесем x под знак дифференциала xdx = (1/2) dx² = (1/2) d (x²+4), сделаем замену переменной x²+4=y, dx²=dy

1

∫ ((1/2) dx²) / (4+x²) =

0

1

= (1/2) ∫ (1/y) dy

0

1

= (1/2) (lny) =

0

1

= (1/2) (ln (x²+4)) =

0

= (1/2) [ ln (4+1) - ln (4+0) ]=

= (1/2) (ln5-ln4)

= (1/2) ln (5/4)

= (1/2) ln1,2