Solve:
$\int \frac{v}{1 + v^{2}} d v$

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Solution

b'Now,
$displaystyleintdfracv1+v2dv$
$=dfrac12displaystyleintdfrac2vdv1+v2$
$=dfrac12displaystyleintdfracd(1+v2)1+v2$
$=dfrac12log|(1+v2)|+c$ . [ Where $c$ is integrating constant]
'

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Similar questions

\int \frac{1 + V}{1 - 2 V - V^{2}} d V =

\int \frac{1 - V}{1 + V^{2}} d v .

\int ⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{log \frac{1}{v}}{v^{2}} ⎞ ⎟ ⎟ ⎟ ⎠ d v

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

u

v = 32 c m, f = 16 c m

\Rightarrow \int d x = \int (2 + \frac{1}{v + 2}) d v

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