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

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

Popular >

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int \frac{10}{( x ^{2} + 9 ) ( x - 1 )} d x

g (x) = tan (2 x) \cdot sin (2 x)

(2 y^{2} x - y) d x + x d y = 0

y = 2 x - x^{2}

\frac{\partial}{\partial x} (\frac{10000 x}{y ^{2}})

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