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

Take the constant out:

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

= - (- \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

y = - x^{3} + 9 x^{2} - 6 x + 9, (7, 65)

y^{^{'}} = \frac{4 x ^{3} + 1}{2 y - 6}

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

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