What is the integral of 1/(2x+4) ?

The integral of 1/(2x+4) is 1/2 ln|2x+4|+C

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integral of 1/(2x+4)

\int \frac{1}{2 x + 4} d x

\frac{1}{2} ln ∣ 2 x + 4 ∣ + C

Solution steps

\int \frac{1}{2 x + 4} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 4

= \frac{1}{2} ln ∣ 2 x + 4 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 4 ∣ + C

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

\int \frac{27}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d y}{d x} = ln (x^{y})

\int (x^{4} - 4^{x} + 4 x - \frac{4}{x}) d x

y = \frac{( x + 7 )}{( x - 5 )}