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

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

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integral of (-1)/(x+1)

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

- ln ∣ x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

= - ln ∣ x + 1 ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ + C

f (x) = x^{2} - 2 x - 3

\frac{\partial}{\partial x} (x^{2} e^{x + y} + 2 y cos (x y) - x y^{2} sin (x y))

x^{2}, \frac{2}{( x ^{2} + 1 )}

\frac{\partial}{\partial x} (cos (x + 7 y))

\frac{1}{x} \frac{d y}{d x} + 2 y - e^{- x^{2}} = 0