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

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

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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

5 + x

\int \frac{3}{4 x ^{2} - 9} d x

\int 2 x e^{3 x} d x

\frac{d}{d x} (cos^{2} (2 x - 1))

f (t) = t^{5}

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