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

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

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integral of 1/(4+y^2)

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

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

Solution steps

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

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{y}{2}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{x ^{2} + 4 x + 3}{x ^{2} + x})

\frac{d}{d x} (2 sec (3 x))

y^{^{''}} + 9 x = - sin (3 x)

\frac{d}{d x} (\frac{1}{16} x^{4} + \frac{1}{2 x ^{2}})

x \to 3 lim (2 x + 3)

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