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

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

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

\frac{2}{- x + 1} + ln ∣ - x + 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

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

Simplify

\frac{1}{1 - x} + ln ∣ 1 - x ∣ + \frac{1}{1 - x} : \frac{2}{- x + 1} + ln ∣ - x + 1 ∣

= \frac{2}{- x + 1} + ln ∣ - x + 1 ∣

Add a constant to the solution

= \frac{2}{- x + 1} + ln ∣ - x + 1 ∣ + C

\int \frac{1}{π} d x

\int \frac{x ^{2} + 9}{x ^{6}} d x

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

\int \frac{1 - x}{4 - x ^{2}} d x

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

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