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

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

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

integral

\frac{( 1 + a )}{( 1 - a )}

- 2 ln ∣ 1 - a ∣ - a + C

Solution steps

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

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

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

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 - a} d a - \int 1 d a

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

\int 1 d a = a

= - 2 ln ∣ 1 - a ∣ - a

Add a constant to the solution

= - 2 ln ∣ 1 - a ∣ - a + C

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

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

\frac{1}{2 x ^{2} y}

x^{4} e^{x}

(1 + x)^{\frac{1}{4}}

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