What is the integral of (2x)/(1-x) ?

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

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

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

- 2 (- 1 + x + ln ∣ 1 - x ∣) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int - \frac{- u + 1}{u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

= 2 (- \int - 1 + \frac{1}{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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (- (- u + ln ∣ u ∣))

Substitute back

u = 1 - x

= 2 (- (- (1 - x) + ln ∣ 1 - x ∣))

Simplify

= - 2 (- 1 + x + ln ∣ 1 - x ∣)

Add a constant to the solution

= - 2 (- 1 + x + ln ∣ 1 - x ∣) + C

\int \frac{- 3 x}{( x - 1 ) ^{2} + 4} d x

y^{^{'''}} - y^{^{''}} + y^{^{'}} + 3 y = 0

\frac{d y}{d t} = 3 y + y^{4}, y (0) = 1

\frac{d}{d x} (- 20 x^{3} + 36 x^{2} - 12 x - 4)

\frac{d ^{2}}{d x ^{2}} (9 z^{2} e^{z})

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