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

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

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integral of 1/(z^2+4)

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

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

Solution steps

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

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{z}{2}

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

Add a constant to the solution

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

y = x, y = 12 x, x = 16

y^{^{''}} + y = cot (x)

\frac{\partial}{\partial y} (ln (3 + x^{2} y^{2}))

\frac{d}{d x} (\frac{sin ( x )}{1 + 2 cos ( x )})

y = e^{x} cos (x)

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