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Verified Answer :

syed514

i think you have to first separate the integral:1/(1+v^2) + v/(1+v^2),
so the integral of the first term is ArcTan (v) and for the integral of the second term i recommend you to do a change of variable:

y= 1+v^2
 so
 dy= 2v
 and
v= dy/2and then you substitute:v/(1+v^2) = (1/2)(dy/y)
and the integral is
 (1/2) (In y)finally you plug in the initial variables:

(1/2)(In [1+v^2])

so the total integral is:

ArcTan (y) + (1/2)(In [1+v^2])

Answer:

\(tan^-1( v)  - \frac{1}{2} ln(1+{v}^2) +c\) is the answer.

Step-by-step explanation:

∫\(\frac{1-v}{1+{v}^2} \\\frac{1}{1+{v}^2}- \frac{v}{1+{v}^2}}\)

∫\(\frac{dv}{1+{v}^2}\)-∫\(\frac{2v}{1+{v}^2}\)dv

\({tan}^{-1} (v)-\frac{1}{2} ln(1+{v}^2) +c\)

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