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

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

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integral of-1/(x+2)

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

- ln ∣ x + 2 ∣ + C

Solution steps

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

Take the constant out:

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

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

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 = x + 2

= - ln ∣ x + 2 ∣

Add a constant to the solution

= - ln ∣ x + 2 ∣ + C

\frac{d y}{d x} = (\frac{( y + 3 )}{( 2 x + 5 )})^{2}

\frac{d y}{d x} = - y^{2} + y

\frac{d}{d x} (x^{n})

10 x^{2} sin (8 x)

\frac{d}{d x} (4 x + 10)