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

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

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

\int \frac{1}{( z + \frac{1}{2} )} d z

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = z + \frac{1}{2}

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

Add a constant to the solution

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

\int \frac{cos ^{5} ( 20 )}{sin ( 20 )} d^{0} dd

x \to 0 lim (\frac{x}{arctan ( 6 x )})

x \to 0 + lim (x^{2} + 3 x - 4)

\frac{d}{d x} (\frac{10}{1 + e ^{- x}})

\frac{d y}{d x} + (\frac{1}{200 + 3 x}) y = 8

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