Question

Integrate:

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

Answer 1

$\int \frac{1-v}{1+v^2}dv$

let $v=tan\theta$ ___(1)

$dv=se{c}^2\theta d\theta$

$\int \frac{1-tan\theta }{1+ta{n}^2\theta }\times se{c}^2\theta d\theta$

$\int \frac{1-tan\theta }{se{c}^2\theta }\times se{c}^2\theta d\theta$

$\int d\theta -\int tan\theta d\theta$

$\theta -(-ln(cos\theta \left)\right)+c$ $[\because \int tan\theta d\theta =lntan\theta ]$

$\theta +ln\left(cos\theta \right)+c$___(2)

from (1)

$\theta =ta{n}^{-1}v$

putting in (1)

$ta{n}^{-1}v+ln\left(ta{n}^{-1}v\right)+c$ Ans