What is the integral of ((y-2)/(y^2-4y)) ?

The integral of ((y-2)/(y^2-4y)) is 1/2 ln|y|+1/2 ln|y-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((y-2)/(y^2-4y))

\int (\frac{y - 2}{y ^{2} - 4 y}) d y

\frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣ + C

Solution steps

\int \frac{y - 2}{y ^{2} - 4 y} d y

Expand

\frac{y - 2}{y ^{2} - 4 y} : \frac{y}{y ^{2} - 4 y} - \frac{2}{y ^{2} - 4 y}

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{y}{y ^{2} - 4 y} d y - \int \frac{2}{y ^{2} - 4 y} d y

\int \frac{y}{y ^{2} - 4 y} d y = ln ∣ y - 4 ∣

\int \frac{2}{y ^{2} - 4 y} d y = 2 (- \frac{1}{4} ln ∣ y ∣ + \frac{1}{4} ln ∣ y - 4 ∣)

= ln ∣ y - 4 ∣ - 2 (- \frac{1}{4} ln ∣ y ∣ + \frac{1}{4} ln ∣ y - 4 ∣)

Simplify

ln ∣ y - 4 ∣ - 2 (- \frac{1}{4} ln ∣ y ∣ + \frac{1}{4} ln ∣ y - 4 ∣) : \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣

= \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ y ∣ + \frac{1}{2} ln ∣ y - 4 ∣ + C

\int tan (\frac{x}{2}) sec (\frac{x}{2}) d x

\int x e^{x^{2}} + \frac{1}{x ln ( x )} d x

\int e^{3 x + 3} d x

\int \frac{1}{x 36 x ^{2} - 25} d x

\int \frac{3 + cos ^{3} ( x )}{cos ^{2} ( x )} d x

What is the integral of ((y-2)/(y^2-4y)) ?
The integral of ((y-2)/(y^2-4y)) is 1/2 ln|y|+1/2 ln|y-4|+C