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

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

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

integral

\frac{- x}{( 1 + x )}

- 1 - x + ln ∣ 1 + x ∣ + C

Solution steps

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

Take the constant out:

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

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

= - (1 + x - ln ∣ 1 + x ∣)

Simplify

= - 1 - x + ln ∣ 1 + x ∣

Add a constant to the solution

= - 1 - x + ln ∣ 1 + x ∣ + C

(1 - t^{2})^{3}

\frac{5 x ^{4}}{x ^{5} + 9}

x \cdot lo g_{10} (\frac{1}{x})

\frac{1}{( x ( ln ( x ) ) )}

u^{- 1}

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