Question

plz answer fast....full explanation plz
CV is constant


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Answer 1

${\underline{\sf{Question}}}$

If cv is constant term , then integrate

$\sf \int \limits_{0}^{q} \dfrac{dq}{cv - q}$

${\underline{\sf{Theory}}}$

$1) \sf \int \dfrac{1}{x} = log(x) + c$

$2) \sf \int \: x {}^{n} dx = \dfrac{x {}^{n + 1} }{ n+ 1} + c$

${\underline{\sf{Solution}}}$

$\sf \int \limits_{0}^{q} \dfrac{dq}{cv - q}$

$= \sf \int \limits_{0}^{q} - 1 \times \dfrac{dq}{q - cv}$

$= - \sf \int \limits_{0}^{q} \dfrac{dq}{q - cv}$

$= \sf - [ log(q - cv)] {}^{q}_{0}$

$\sf = - [ \log(q - cv) - \log( - cv)]$

We know that

$\log \: a - \log \: b = \log \frac{a}{b}$

$\sf = - [ \log (\dfrac{q - cv}{ - cv}) ]$

$\sf = \log( \dfrac{ - cv}{q - cv} )$

It is the required solution!