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

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

5 y^{^{'}} = 2 y

\int 2 x - 2 d x

\int x - 2 y + 2

x \to \infty lim (- 2 + x)

\frac{\partial}{\partial x} (\frac{6 x - 9 y}{6 x + 9 y})