Question

SOLVE. integration of (1-v) /(1+v^2)

Answer 1

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 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

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