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

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

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integral of 1/(1+x/2)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 ln ∣ u ∣

Substitute back

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{3 ^{n}}{4 ^{n + 1}}

x \to 1 lim (4 x - 1)

\frac{d}{d x} (\frac{( - ln ( 1 + 6 x ) )}{x})

\frac{150000}{( 4 t + 75 ) ^{2}}

\int \frac{1}{6 x} d x