What is the integral of 1/((x-3)ln(x-3)) ?

The integral of 1/((x-3)ln(x-3)) is ln|ln(x-3)|+C

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integral of 1/((x-3)ln(x-3))

\int \frac{1}{( x - 3 ) ln ( x - 3 )} d x

ln ∣ ln (x - 3) ∣ + C

Solution steps

\int \frac{1}{( x - 3 ) ln ( x - 3 )} 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 - 3) ∣

Add a constant to the solution

= ln ∣ ln (x - 3) ∣ + C

\frac{d x}{d t} = x - x^{3}

(x e^{- 2 x})^{^{'}}

\int tan^{3} (π x) d x

\int_{0}^{\infty} e^{- 2 a x} d x

x \to 9 + lim (\frac{x + 8}{x + 9})

What is the integral of 1/((x-3)ln(x-3)) ?
The integral of 1/((x-3)ln(x-3)) is ln|ln(x-3)|+C