What is the integral of (ln(x))/(x(6-ln(x))) ?

The integral of (ln(x))/(x(6-ln(x))) is 6-ln(x)-6ln|6-ln(x)|+C

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integral of (ln(x))/(x(6-ln(x)))

\int \frac{ln ( x )}{x ( 6 - ln ( x ) )} d x

6 - ln (x) - 6 ln ∣ 6 - ln (x) ∣ + C

Solution steps

\int \frac{ln ( x )}{x ( 6 - ln ( x ) )} d x

Apply u-substitution

: \int - \frac{- u + 6}{u} d u

= \int - \frac{- u + 6}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{- u + 6}{u} d u

Expand

\frac{- u + 6}{u} : - 1 + \frac{6}{u}

= - \int - 1 + \frac{6}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int 1 d u + \int \frac{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= - (- u + 6 ln ∣ u ∣)

Substitute back

u = 6 - ln (x)

= - (- (6 - ln (x)) + 6 ln ∣ 6 - ln (x) ∣)

Simplify

= 6 - ln (x) - 6 ln ∣ 6 - ln (x) ∣

Add a constant to the solution

= 6 - ln (x) - 6 ln ∣ 6 - ln (x) ∣ + C

\int y e^{x y} - \frac{1}{y} d x

\int x \cdot x^{2} + 9 d x

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

\int 1 - (sin (x))^{2} d x

What is the integral of (ln(x))/(x(6-ln(x))) ?
The integral of (ln(x))/(x(6-ln(x))) is 6-ln(x)-6ln|6-ln(x)|+C