Question

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

Answer 1

$\displaystyle\int\frac{1-v}{1+v^2}\,\mathrm dv=\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv$

The first integral is already standard and has an antiderivative in terms of $\arctan v$. For the second integral, take $w=1+v^2$ so that $\dfrac{\mathrm dw}2=v\,\mathrm dv$. Then

$\displaystyle\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv=\arctan v-\frac12\int\frac{\mathrm dw}w$
$=\arctan v-\dfrac12\ln|w|+C$
$=\arctan v-\dfrac12\ln(1+v^2)+C$