What is the integral of 1/((y-2)) ?

The integral of 1/((y-2)) is ln|y-2|+C

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integral of 1/((y-2))

integral

\frac{1}{( y - 2 )}

ln ∣ y - 2 ∣ + C

Solution steps

\int \frac{1}{y - 2} d y

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 = y - 2

= ln ∣ y - 2 ∣

Add a constant to the solution

= ln ∣ y - 2 ∣ + C

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

(4 x + 9)^{\frac{1}{2}}

(2 x + 8)^{3}

x^{2} arcsin (x)

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