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

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

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

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

Solution steps

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

Expand

\frac{x + ln ( 1 - x )}{x ^{2}} : \frac{1}{x} + \frac{ln ( 1 - x )}{x ^{2}}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Simplify

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

Add a constant to the solution

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

\int (\frac{3}{x ^{2}} + 4 - \frac{1}{4 x ^{3}}) d x

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

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

\int x^{2} sec (x^{3}) tan (x^{3}) d x

4 x^{2}

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