Question

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

Answer 1

`\displaystyle\int(1-v)/(1+v^2)\,\mathrm dv=\int(\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
`\frac{\mathrm dw}2=v\,\mathrm dv`. Then


`\displaystyle\int(\mathrm dv)/(1+v^2)-\int\frac v{1+v^2}\,\mathrm dv=\arctan v-\frac12\int\frac{\mathrm dw}w`

`=\arctan v-\frac12\ln|w|+C`

`=\arctan v-\frac12\ln(1+v^2)+C`