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))

integral

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

ln ∣ x - 1 ∣ + C

Solution steps

\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

x^{3} (e^{x})^{2}

\frac{1}{( x ^{3} + 1 )}

\frac{( x - 1 )}{( x + 1 ) ^{2}}

- x^{- 2}

\frac{r}{( 1 + r ^{2} ) ^{2}}