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

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

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

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

- arctan (v) + v + C

Solution steps

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

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

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

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 1 d v

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

\int 1 d v = v

= - arctan (v) + v

Add a constant to the solution

= - arctan (v) + v + C

\int \frac{1}{x x ^{2} - 400} d x

\int \frac{x ^{2} + 4}{x ^{3}} d x

x^{3} e^{x^{2}}

\int \frac{6 x ^{2} + 5 x + 4}{( 1 + x ) ( 1 + 4 x ^{2} )} d x

\int \frac{2 x ^{3} + 5 x}{x ^{4} + 5 x ^{2} + 4} d x

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