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

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

Popular >

integral of-x/(x+1)

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

- x - 1 + ln ∣ x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

Substitute back

u = x + 1

= - (x + 1 - ln ∣ x + 1 ∣)

Simplify

= - x - 1 + ln ∣ x + 1 ∣

Add a constant to the solution

= - x - 1 + ln ∣ x + 1 ∣ + C

f (x) = \frac{1}{x ^{6}}, (- 2, \frac{1}{64})

\int_{0}^{1} \frac{4}{x ^{2} + 3 x + 2} d x

\frac{d}{d x} (x^{\frac{2}{5}} (x - 5))

y^{^{''}} - 6 y^{^{'}} = 36 t, y (0) = 6, y^{^{'}} (0) = 0

\int_{0}^{8} x^{3} e^{x} d x