What is the integral of (y-2)/(y(y-4)(y+3)) ?

The integral of (y-2)/(y(y-4)(y+3)) is 1/14 ln|y-4|-5/21 ln|y+3|+1/6 ln|y|+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(y-4)(y+3))

\int \frac{y - 2}{y ( y - 4 ) ( y + 3 )} d y

\frac{1}{14} ln ∣ y - 4 ∣ - \frac{5}{21} ln ∣ y + 3 ∣ + \frac{1}{6} ln ∣ y ∣ + C

Solution steps

\int \frac{y - 2}{y ( y - 4 ) ( y + 3 )} d y

Expand

\frac{y - 2}{y ( y - 4 ) ( y + 3 )} : \frac{y}{y ( y - 4 ) ( y + 3 )} - \frac{2}{y ( y - 4 ) ( y + 3 )}

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 ( y - 4 ) ( y + 3 )} d y - \int \frac{2}{y ( y - 4 ) ( y + 3 )} d y

\int \frac{y}{y ( y - 4 ) ( y + 3 )} d y = \frac{1}{7} ln ∣ y - 4 ∣ - \frac{1}{7} ln ∣ y + 3 ∣

\int \frac{2}{y ( y - 4 ) ( y + 3 )} d y = 2 (\frac{1}{21} ln ∣ y + 3 ∣ - \frac{1}{12} ln ∣ y ∣ + \frac{1}{28} ln ∣ y - 4 ∣)

= \frac{1}{7} ln ∣ y - 4 ∣ - \frac{1}{7} ln ∣ y + 3 ∣ - 2 (\frac{1}{21} ln ∣ y + 3 ∣ - \frac{1}{12} ln ∣ y ∣ + \frac{1}{28} ln ∣ y - 4 ∣)

Simplify

\frac{1}{7} ln ∣ y - 4 ∣ - \frac{1}{7} ln ∣ y + 3 ∣ - 2 (\frac{1}{21} ln ∣ y + 3 ∣ - \frac{1}{12} ln ∣ y ∣ + \frac{1}{28} ln ∣ y - 4 ∣) : \frac{1}{14} ln ∣ y - 4 ∣ - \frac{5}{21} ln ∣ y + 3 ∣ + \frac{1}{6} ln ∣ y ∣

= \frac{1}{14} ln ∣ y - 4 ∣ - \frac{5}{21} ln ∣ y + 3 ∣ + \frac{1}{6} ln ∣ y ∣

Add a constant to the solution

= \frac{1}{14} ln ∣ y - 4 ∣ - \frac{5}{21} ln ∣ y + 3 ∣ + \frac{1}{6} ln ∣ y ∣ + C

\frac{( x ^{2} - 4 )}{x}

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

\int \frac{2}{4 + x ^{2}} d x

y = \frac{1}{x}, y = x, y = 2

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

What is the integral of (y-2)/(y(y-4)(y+3)) ?
The integral of (y-2)/(y(y-4)(y+3)) is 1/14 ln|y-4|-5/21 ln|y+3|+1/6 ln|y|+C