${\rm{ Put }}1 + \sqrt x = t$
$ \Rightarrow \frac{1}{{2\sqrt x }}dx = dt$
$\Rightarrow \quad I=\int \frac{2 d t}{t \sqrt{2 t-t^{2}}}$
Again put $t=\frac{1}{z}$
$ \Rightarrow d t=\frac{-1}{2^{2}} d z$
$ = I = 2\int {\frac{{ - \frac{1}{{{z^2}}}dz}}{{\frac{1}{z}\sqrt {\frac{2}{z} - \frac{1}{{{z^2}}}} }}} $
$= 2\int {\frac{{ - dz}}{{\sqrt {2z - 1} }}} $
$=-2 \sqrt{2 z-1}+c$
$=-2 \sqrt{\frac{2}{t}-1+c}$
$=-2 \sqrt{\frac{2-t}{t}}+c$
$=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$
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$\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1, x \in[2,4]$ જ્યાં $f(2)=\frac{1}{2}$ અને $f(4)=\frac{1}{4}$ છે.
નીચેના બે વિધાનો ધ્યાને લો.
$(A)$ : પ્રત્યેક $x \in[2,4]$ માટે. $f(x) \leq 1$
$(B)$ : પ્રત્યેક $x \in[2,4]$ માટ $f(x) \geq \frac{1}{8}$ તો,