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

The integral of z/(z^2+1) is 1/2 ln|z^2+1|+C

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integral of z/(z^2+1)

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

\frac{1}{2} ln z^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} 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} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = z^{2} + 1

= \frac{1}{2} ln z^{2} + 1

Add a constant to the solution

= \frac{1}{2} ln z^{2} + 1 + C

\frac{d}{d t} (arctan (t))

y = (x + 1)^{2}

\frac{d y}{d x} = - x e^{y}

\frac{\partial}{\partial x} (8 sin (x y))

\frac{d}{d y} (\frac{1}{4 y ^{2}})

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