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

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

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

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

ln ∣ x ln (x) - x ∣ + 3 ln ∣ ln (x) - 1 ∣ + C

Solution steps

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

Expand

x (ln (x) - 1) : x ln (x) - x

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

Expand

\frac{ln ( x ) + 3}{x ln ( x ) - x} : \frac{ln ( x )}{x ln ( x ) - x} + \frac{3}{x ln ( x ) - x}

Apply the Sum Rule:

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

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

\int \frac{ln ( x )}{x ln ( x ) - x} d x = ln ∣ x ln (x) - x ∣

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

= ln ∣ x ln (x) - x ∣ + 3 ln ∣ ln (x) - 1 ∣

Add a constant to the solution

= ln ∣ x ln (x) - x ∣ + 3 ln ∣ ln (x) - 1 ∣ + C

sin (t - 5)

y = (3 x^{2} + 3 x - 1)^{4}

(x^{2} - 3 y^{2}) d x + (2 x y) d y = 0

\frac{d}{d x} (\frac{x}{x})

f (x) = 4 x^{2} - 4 x + 1, (- 1, 8)

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