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

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

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integral of 1/((r-1))

integral

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

ln ∣ r - 1 ∣ + C

Solution steps

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

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 = r - 1

= ln ∣ r - 1 ∣

Add a constant to the solution

= ln ∣ r - 1 ∣ + C

\frac{cos ( 1 )}{x}

\frac{1}{( sin ^{2} ( x ) \cdot cos ^{4} ( x ) )}

\frac{x ^{4}}{( x + 1 )}

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

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