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Popular >

integral of z/(z+1)

Solution

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

Solution

z + 1 - ln ∣ z + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Substitute back

u = z + 1

= z + 1 - ln ∣ z + 1 ∣

Add a constant to the solution

= z + 1 - ln ∣ z + 1 ∣ + C

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4 x^{2} y^{^{'}} = y^{^{'}} + 4 x e^{- y}

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2 sec (2 t) 2 t \cdot y^{^{'}} = 1

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y = 3 x^{3}, y = 75 x

Frequently Asked Questions (FAQ)

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