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)

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

\frac{d}{d x} (9 x^{l n (9 x)})

g (x) = - x^{5} + 4 x^{2} - 2 x + 2

\int \frac{sin ( 2 x )}{18 + cos ^{2} ( x )} d x

\frac{d}{d x} (\frac{4 x ^{5} - 5 x ^{4} + 6 x - 2}{x ^{4}})

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