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Popular >

integral of 1/(z+1)

Solution

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

Solution

ln ∣ z + 1 ∣ + C

Solution steps

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

= ln ∣ z + 1 ∣

Add a constant to the solution

= ln ∣ z + 1 ∣ + C

Popular Examples

derivative of-(x^2)

\frac{d}{d x} (- (x^{2}))

derivative of (-11/(sqrt(x)))

\frac{d}{d x} (\frac{- 11}{x})

limit as x approaches 0+of ((sin(x))/x)^{1/x}

x \to 0 + lim ((\frac{sin ( x )}{x})^{\frac{1}{x}})

derivative of (1+ln(x)/(x^2-ln(x)))

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

integral of 1/2 x^2ln(x)-(x^2)/4-(x^2)/2

\int \frac{1}{2} x^{2} ln (x) - \frac{x ^{2}}{4} - \frac{x ^{2}}{2} d x

Frequently Asked Questions (FAQ)

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