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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{ln ( x )}{x - 1} - ln ∣ x ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

= - \frac{ln ( x )}{x - 1} - ln ∣ x ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - \frac{ln ( x )}{x - 1} - ln ∣ x ∣ + ln ∣ x - 1 ∣ + C

y = 12 x, (1, 12)

\int sin (x) d x

(t - 3)^{2}

n \to \infty lim (5^{- n})

\int 3 x^{2} - 1 d x

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