What is the integral of 3/(-1-x) ?

The integral of 3/(-1-x) is -3ln|1+x|+C

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integral of 3/(-1-x)

\int \frac{3}{- 1 - x} d x

- 3 ln ∣ 1 + x ∣ + C

Solution steps

\int \frac{3}{- 1 - x} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{- 1 - x} d x

Apply u-substitution

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

= 3 \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= 3 (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- ln ∣ u ∣)

Substitute back

u = - 1 - x

= 3 (- ln ∣ - 1 - x ∣)

Simplify

3 (- ln ∣ - 1 - x ∣) : - 3 ln ∣ 1 + x ∣

= - 3 ln ∣ 1 + x ∣

Add a constant to the solution

= - 3 ln ∣ 1 + x ∣ + C

x \to 0 lim (\frac{1}{2} + x^{2})

\int 9 x^{2} sin (π x) d x

\frac{d}{d t} (3 t^{3})

\frac{\partial}{\partial x} (\frac{x - y}{x + y})

y = \frac{4}{3} π r^{2}