Question Number 216093 by mnjuly1970 last updated on 27/Jan/25
Answered by mahdipoor last updated on 27/Jan/25
$${note}\::\:\mathrm{0}\leqslant{x}<\mathrm{1}\:\:,\:\:{y}^{\mathrm{2}} =\left(\mathrm{1}+{y}^{\mathrm{2}} \right){x} \\ $$$${get}\:\:{f}={ln}\left({y}\right)+{ln}\left({y}'\right) \\ $$$$\Rightarrow\frac{{df}}{{dx}}=\frac{{y}^{'} }{{y}}+\frac{{y}^{''} }{{y}'}=\mathrm{2}\varphi \\ $$$${f}={ln}\left({yy}^{'} \right)={ln}\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{{d}}{{dx}}\left({y}^{\mathrm{2}} \right)\right)= \\ $$$${ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{ln}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\Rightarrow\frac{{df}}{{dx}}=\frac{\mathrm{2}}{\mathrm{1}−{x}}=\mathrm{2}\varphi\:\Rightarrow\:\varphi=\frac{\mathrm{1}}{\mathrm{1}−{x}}=\mathrm{1}+{y}^{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\:\infty} \:\frac{{ln}\left(\mathrm{1}+{y}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}\:=\:... \\ $$
Commented by mnjuly1970 last updated on 27/Jan/25
$${thank}\:{you}\:{so}\:{much} \\ $$