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

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

integral

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

\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

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

\frac{1}{( cos ^{2} ( x ) ( 1 - tan ( x ) ) ^{2} )}

\frac{7 x + 2}{1 - 25 x ^{2}}

a^{2} (e^{- a})^{2} x^{4}

\frac{ln ( 4 x )}{x ^{4}}

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