What is the integral of ((y^4-8))/(y^3+2y^2) ?

The integral of ((y^4-8))/(y^3+2y^2) is (y^2)/2-2y+2ln|y|+4/y+2ln|y+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((y^4-8))/(y^3+2y^2)

\int \frac{( y ^{4} - 8 )}{y ^{3} + 2 y ^{2}} d y

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

Solution steps

\int \frac{y ^{4} - 8}{y ^{3} + 2 y ^{2}} d y

Take the partial fraction of

\frac{y ^{4} - 8}{y ^{3} + 2 y ^{2}} : y - 2 + \frac{2}{y} - \frac{4}{y ^{2}} + \frac{2}{y + 2}

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

\int y d y = \frac{y ^{2}}{2}

\int 2 d y = 2 y

\int \frac{2}{y} d y = 2 ln ∣ y ∣

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

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

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

Simplify

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

Add a constant to the solution

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

\int csc (1 - 2 x) cot (1 - 2 x) d x

\int \frac{b x + c}{x ^{2} - 8 x + 25} d x

\int \frac{15}{( 0.5 t + 1 ) ^{2}} d t

\int 5 x x - 1 d x

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

What is the integral of ((y^4-8))/(y^3+2y^2) ?
The integral of ((y^4-8))/(y^3+2y^2) is (y^2)/2-2y+2ln|y|+4/y+2ln|y+2|+C