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

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

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integral of 1/((x-2)ln(x-2))

\int \frac{1}{( x - 2 ) ln ( x - 2 )} d x

ln ∣ ln (x - 2) ∣ + C

Solution steps

\int \frac{1}{( x - 2 ) ln ( x - 2 )} d x

Apply u-substitution

: \int \frac{1}{u ln ( u )} d u

= \int \frac{1}{u ln ( u )} d u

Apply u-substitution

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

= \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ ln (x - 2) ∣

Add a constant to the solution

= ln ∣ ln (x - 2) ∣ + C

\int (16 x^{3} - 3 x^{2} + 2) d x

x \to 2 lim (\frac{3 x ^{2} + 4 - 4 x ^{2} + 6 x}{x ^{2} - 4})

\int \frac{cos ( x )}{3 + sin ( x )} d x

\int \frac{x ^{3} + 4 x - 2}{( x - 2 ) ( x + 1 )} d x

f (x) = \frac{x ^{3}}{4}, at x = - 6

What is the integral of 1/((x-2)ln(x-2)) ?
The integral of 1/((x-2)ln(x-2)) is ln|ln(x-2)|+C