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

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

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integral of-1/(x-1)

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

- ln ∣ x - 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{u} d u

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = x - 1

= - ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x - 1 ∣ + C

\int_{3}^{x^{3}} t^{3} d t

\frac{1}{x}

\frac{1}{3} x (x - 3)

2 x^{2} - 5 x + 3, at x = 0

x^{t a n (x)}