What is the integral of (2y-x)/(-(y+xy)) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2y-x)/(-(y+xy))

\int \frac{2 y - x}{- ( y + x y )} d x

\frac{1}{y} + \frac{x}{y} - \frac{1}{y} ln ∣ 1 + x ∣ - 2 ln ∣ 1 + x ∣ + C

Solution steps

\int \frac{2 y - x}{- ( y + x y )} d x

Take the constant out:

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

= - \int \frac{- x + 2 y}{y + y x} d x

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

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

Take the constant out:

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

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

Expand

\frac{- x + 2 y}{1 + x} : - \frac{x}{1 + x} + \frac{2 y}{1 + x}

Apply the Sum Rule:

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

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

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

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

= - \frac{1}{y} (- (1 + x - ln ∣ 1 + x ∣) + 2 y ln ∣ 1 + x ∣)

Simplify

- \frac{1}{y} (- (1 + x - ln ∣ 1 + x ∣) + 2 y ln ∣ 1 + x ∣) : \frac{1}{y} + \frac{x}{y} - \frac{1}{y} ln ∣ 1 + x ∣ - 2 ln ∣ 1 + x ∣

= \frac{1}{y} + \frac{x}{y} - \frac{1}{y} ln ∣ 1 + x ∣ - 2 ln ∣ 1 + x ∣

Add a constant to the solution

= \frac{1}{y} + \frac{x}{y} - \frac{1}{y} ln ∣ 1 + x ∣ - 2 ln ∣ 1 + x ∣ + C

\int 5 ln (x^{2} + 5) d x

f (x) = (1 + 2 x)^{12}, at x = 0

4 y^{^{''}} + 4 y^{^{'}} + 17 y = 0

2 y^{^{'}} + y = 3 t

\frac{\partial}{\partial x} (10 x + 5 y)

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