What is the integral of 1/(4+v^2) ?

The integral of 1/(4+v^2) is 1/2 arctan(v/2)+C

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integral of 1/(4+v^2)

\int \frac{1}{4 + v ^{2}} d v

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

Solution steps

\int \frac{1}{4 + v ^{2}} d v

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= \frac{1}{2} arctan (u)

Substitute back

u = \frac{v}{2}

= \frac{1}{2} arctan (\frac{v}{2})

Add a constant to the solution

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

f (x) = \frac{1 - x ^{2}}{x}

y = 10 csc (5 x^{5} - 4 x + 4)

\int \frac{3 x ^{2}}{( 1 - x ) ( x ^{2} + x + 1 )} d x

x \to 4 lim (- x^{2} + 6 x - 3)

n = 0 \sum \infty x^{2 n + 1}

What is the integral of 1/(4+v^2) ?
The integral of 1/(4+v^2) is 1/2 arctan(v/2)+C