What is the integral of 1/((x-4)ln(x-4)) ?

The integral of 1/((x-4)ln(x-4)) is ln|ln(x-4)|+C

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integral of 1/((x-4)ln(x-4))

\int \frac{1}{( x - 4 ) ln ( x - 4 )} d x

ln ∣ ln (x - 4) ∣ + C

Solution steps

\int \frac{1}{( x - 4 ) ln ( x - 4 )} 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 - 4) ∣

Add a constant to the solution

= ln ∣ ln (x - 4) ∣ + C

y^{^{''}} + 2 y^{^{'}} - 8 y = 0

\int \frac{cos ( x )}{6 + sin ( x )} d x

\frac{6}{s ( s + 4 )}

\frac{d}{d x} (a x^{2} + b x + c)

(x^{2} + 1) \frac{d y}{d x} + 2 x y = 4 x^{2}

What is the integral of 1/((x-4)ln(x-4)) ?
The integral of 1/((x-4)ln(x-4)) is ln|ln(x-4)|+C