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

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

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integral of 1/(a+v^2)

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

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

Solution steps

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

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{1}{a} v

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

Simplify

\frac{1}{a} a arctan (\frac{1}{a} v) : \frac{1}{a} arctan (\frac{1}{a} v)

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

Add a constant to the solution

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

\frac{d}{d x} (3 x^{2} - 18 x)

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