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

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

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

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

- ln 1 - z^{2} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int \frac{z}{1 - z ^{2}} d z

Apply u-substitution

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

= 2 \cdot \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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 1 - z^{2}

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

Simplify

2 (- \frac{1}{2} ln 1 - z^{2}) : - ln 1 - z^{2}

= - ln 1 - z^{2}

Add a constant to the solution

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

x \to 0 + lim (\frac{4}{x ^{2}})

\int 2 x y^{3} z^{4} d x

\int \frac{y - 2}{y + 3} d y

\frac{\partial}{\partial s} (s^{2} t^{2})

\frac{d}{d x} (\frac{x}{1 + 4 x ^{2}})

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