Question-216093

Question Number 216093 by mnjuly1970 last updated on 27/Jan/25

Answered by mahdipoor last updated on 27/Jan/25

n o t e : 0 ⩽ x < 1, y^{2} = (1 + y^{2}) x

g e t f = l n (y) + l n (y^{'})

\Rightarrow \frac{d f}{d x} = \frac{y^{^{'}}}{y} + \frac{y^{^{″}}}{y^{'}} = 2 φ

f = l n ({y y}^{^{'}}) = l n (\frac{1}{2} \times \frac{d}{d x} (y^{2})) =

l n (\frac{1}{2}) + l n (\frac{1}{{(1 - x)}^{2}})

\Rightarrow\Rightarrow \frac{d f}{d x} = \frac{2}{1 - x} = 2 φ \Rightarrow φ = \frac{1}{1 - x} = 1 + y^{2}

\Rightarrow \int_{0}^{\infty} \frac{l n (1 + y)}{1 + y^{2}} d y = \dots

Commented by mnjuly1970 last updated on 27/Jan/25

t h a n k y o u s o m u c h