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

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

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integral of (1+z)/(1-2z-z^2)

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

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

Solution steps

\int \frac{1 + z}{1 - 2 z - z ^{2}} 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 = 1 - 2 z - z^{2}

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

Add a constant to the solution

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

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