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

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

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

integral

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

- 1 - x + 2 ln ∣ 1 + x ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{- u + 2}{u} d u

= \int \frac{- u + 2}{u} d u

Expand

\frac{- u + 2}{u} : - 1 + \frac{2}{u}

= \int - 1 + \frac{2}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int \frac{2}{u} d u

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= - u + 2 ln ∣ u ∣

Substitute back

u = 1 + x

= - (1 + x) + 2 ln ∣ 1 + x ∣

Simplify

= - 1 - x + 2 ln ∣ 1 + x ∣

Add a constant to the solution

= - 1 - x + 2 ln ∣ 1 + x ∣ + C

\frac{1}{( 2 + sin ( 3 x ) )}

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

x^{3} sin (x^{2})

e^{\frac{x}{3}}

\frac{x ^{2}}{( x + 3 ) ^{3}}

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