[tex]\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[/tex]
The first integral is already standard and has an antiderivative in terms of [tex]\arctan v[/tex]. For the second integral, take [tex]w=1+v^2[/tex] so that [tex]\dfrac{\mathrm dw}2=v\,\mathrm dv[/tex]. Then
[tex]\displaystyle\int\frac{\mathrm dv}{1+v^2}-\int\frac v{1+v^2}\,\mathrm dv=\arctan v-\frac12\int\frac{\mathrm dw}w[/tex]
[tex]=\arctan v-\dfrac12\ln|w|+C[/tex]
[tex]=\arctan v-\dfrac12\ln(1+v^2)+C[/tex]
