What is the integral of z/(2z^2+8) ?

The integral of z/(2z^2+8) is 1/4 ln|2z^2+8|+C

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integral of z/(2z^2+8)

\int \frac{z}{2 z ^{2} + 8} d z

\frac{1}{4} ln 2 z^{2} + 8 + C

Solution steps

\int \frac{z}{2 z ^{2} + 8} d z

Apply u-substitution

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 z^{2} + 8

= \frac{1}{4} ln 2 z^{2} + 8

Add a constant to the solution

= \frac{1}{4} ln 2 z^{2} + 8 + C

What is the integral of z/(2z^2+8) ?
The integral of z/(2z^2+8) is 1/4 ln|2z^2+8|+C