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

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

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integral of z/(1+z^2)

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

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

Solution steps

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

Apply u-substitution

= \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 = 1 + z^{2}

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

Add a constant to the solution

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

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