What is the integral of 1/(1+\frac{2z){L}} ?

The integral of 1/(1+\frac{2z){L}} is L/2 ln|1+(2z)/L |+C

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integral of 1/(1+\frac{2z){L}}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{L}{2 u} d u

= \int \frac{L}{2 u} d u

Take the constant out:

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

= \frac{L}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{L}{2} ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} + 6 x + 25} d x

\frac{1}{x + 1}

\int sin^{2} (\frac{π x}{L}) d x

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

What is the integral of 1/(1+\frac{2z){L}} ?
The integral of 1/(1+\frac{2z){L}} is L/2 ln|1+(2z)/L |+C