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

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

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

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

ln z^{2} + 1 + C

Solution steps

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = z^{2} + 1

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

Simplify

2 \cdot \frac{1}{2} ln z^{2} + 1 : ln z^{2} + 1

= ln z^{2} + 1

Add a constant to the solution

= ln z^{2} + 1 + C

\frac{\partial}{\partial s} (ln (r s^{8} t^{8}) + 1)

cos (sin (e^{x}))

\int_{1}^{2} \frac{2 x - 1}{x} d x

t (\frac{d y}{d t}) + 5 y = 4 t, y (1) = 2

x^{- \frac{3}{2}}

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