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

\int sin^{3} (θ) cos^{4} (θ) d θ

\int \frac{4}{x ln ^{3} ( x )} d x

e^{- (t - 3)}

x \to 0 + lim (\frac{4}{5 x ^{\frac{1}{3}}})

\frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}

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