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

Answered by mahdipoor last updated on 27/Jan/25

$$note:0⩽x<1,y^2=(1+y^2)x$$$$getf=ln\left(y\right)+ln\left(y^{\prime }\right)$$$$\Rightarrow \frac{df}{dx}=\frac{y^{{}^{\prime }}}{y}+\frac{y^{{}^{″}}}{y^{\prime }}=2\phi$$$$f=ln\left({yy}^{{}^{\prime }}\right)=ln\left(\frac{1}{2}\times \frac{d}{dx}\left(y^2\right)\right)=$$$$ln\left(\frac{1}{2}\right)+ln\left(\frac{1}{{(1-x)}^2}\right)$$$$\Rightarrow \Rightarrow \frac{df}{dx}=\frac{2}{1-x}=2\phi \Rightarrow \phi =\frac{1}{1-x}=1+y^2$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{ln(1+y)}{1+y^2}dy=...$$

Commented by mnjuly1970 last updated on 27/Jan/25

$$thankyousomuch$$

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