E is inversely proportional to r^2 then how in 1st question of physics doubt class E of left side became 2E in left side ,it should be 4E

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Q. If

\int \frac{2 e^{5 x} + e^{4 x} - 4 e^{3 x} + 4 e^{2 x} + 2 e^{x}}{(e^{2 x} + 4) {(e^{2 x} - 1)}^{2}} d x

= {tan}^{- 1} (e^{x / 2}) - \frac{K}{248 (e^{2 x} - 1)} + C

then K is equal to.

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, E is inversely proportional to

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Q. If

\int (e^{2 x} + 2 e^{x} - e^{- x} - 1) e^{(e^{x} + e^{- x})} d x = g (x) e^{(e^{x} + e^{- x})} + c

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Sir,

I am Shaun job of class 10, I have a doubt in physics on the topic of optics,

This is the question Sir

If the Rayleigh scattering holds good, then the intensity of scattering is inversely proportional to the fourth power of the wavelength, then when sunlight enters the earth's atmosphere, the violet light should get scattered the most and blue light also gets scattered, but less than violet light, so my question is why doesnt the sky appear violet instead of blue??

Thank you Sir

Shaun

Q. During electrolysis,

O_{2} (g)

is evolved at anode in

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

\begin{matrix} I n (d) \end{matrix}

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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