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

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

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integral of z/(z^4+4)

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

\frac{1}{4} arctan (\frac{z ^{2}}{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 4 )} d u

= \int \frac{1}{2 ( u ^{2} + 4 )} 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} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (\frac{z ^{2}}{2}) : \frac{1}{4} arctan (\frac{z ^{2}}{2})

= \frac{1}{4} arctan (\frac{z ^{2}}{2})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{z ^{2}}{2}) + C

y^{^{''}} + 4 y^{^{'}} + 3 = 0

\frac{d y}{d x} = (x + y - 3)^{2}

cos (x), 0

\frac{d}{d x} (x^{3} + \frac{4 x ^{2}}{3} - 2 x + 5)

\int x^{3} x^{2} + 36 d x

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