What is the integral of (1-v)/(1+v^2) ?

The integral of (1-v)/(1+v^2) is arctan(v)-1/2 ln|1+v^2|+C

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integral of (1-v)/(1+v^2)

\int \frac{1 - v}{1 + v ^{2}} d v

arctan (v) - \frac{1}{2} ln 1 + v^{2} + C

Solution steps

\int \frac{1 - v}{1 + v ^{2}} d v

Expand

\frac{1 - v}{1 + v ^{2}} : \frac{1}{1 + v ^{2}} - \frac{v}{1 + v ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{1 + v ^{2}} d v - \int \frac{v}{1 + v ^{2}} d v

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

\int \frac{v}{1 + v ^{2}} d v = \frac{1}{2} ln 1 + v^{2}

= arctan (v) - \frac{1}{2} ln 1 + v^{2}

Add a constant to the solution

= arctan (v) - \frac{1}{2} ln 1 + v^{2} + C

What is the integral of (1-v)/(1+v^2) ?
The integral of (1-v)/(1+v^2) is arctan(v)-1/2 ln|1+v^2|+C