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)

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

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y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = - 2, y^{^{'}} (0) = 4

\frac{d}{d x} (\frac{1}{x + sin ( x )})

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