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

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

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

\int \frac{1 + x}{1 - x} d x

- 2 ln ∣ 1 - x ∣ - x + C

Solution steps

\int \frac{1 + x}{1 - x} d x

\frac{1 + x}{1 - x} = \frac{2}{1 - x} - 1

= \int \frac{2}{1 - x} - 1 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{2}{1 - x} d x - \int 1 d x

\int \frac{2}{1 - x} d x = - 2 ln ∣ 1 - x ∣

\int 1 d x = x

= - 2 ln ∣ 1 - x ∣ - x

Add a constant to the solution

= - 2 ln ∣ 1 - x ∣ - x + C

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

f (x) = a^{2 x}

8 y y^{^{'}} = x

\frac{6}{x ^{5}} - \frac{2}{x}

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