Mathematics

Integrate:
$$\int \frac{1-V}{1+V^{2}}dv.$$


SOLUTION
$$ \int \frac{1-v}{1+v^{2}}dv $$
let $$ v = tan\theta $$ ___(1)
$$ dv = sec^{2}\theta d\theta $$ 
$$ \int \frac{1-tan\theta }{1+tan^{2}\theta }\times sec^{2}\theta d\theta $$
$$ \int \frac{1-tan\theta }{sec^{2}\theta }\times sec^{2}\theta d\theta $$
$$ \int d\theta -\int tan\theta d\theta $$
$$ \theta -(-ln(cos\theta ))+c $$  $$[\because \int tan\theta d\theta = ln\, tan\theta ] $$
$$ \theta +ln(cos\theta )+c$$___(2)
from (1)
$$ \theta = tan^{-1}v $$
putting in (1)
$$ tan^{-1}v+ln(tan^{-1}v)+c $$ Ans 
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Subjective Medium Published on 17th 09, 2020
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