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

The integral of ((y+3))/(y^3+4y^2) is 1/16 ln|y|-3/(4y)-1/16 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+3))/(y^3+4y^2)

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

\frac{1}{16} ln ∣ y ∣ - \frac{3}{4 y} - \frac{1}{16} ln ∣ y + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{y + 3}{y ^{3} + 4 y ^{2}} : \frac{1}{16 y} + \frac{3}{4 y ^{2}} - \frac{1}{16 ( y + 4 )}

= \int \frac{1}{16 y} + \frac{3}{4 y ^{2}} - \frac{1}{16 ( y + 4 )} 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 \frac{1}{16 y} d y + \int \frac{3}{4 y ^{2}} d y - \int \frac{1}{16 ( y + 4 )} d y

\int \frac{1}{16 y} d y = \frac{1}{16} ln ∣ y ∣

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

\int \frac{1}{16 ( y + 4 )} d y = \frac{1}{16} ln ∣ y + 4 ∣

= \frac{1}{16} ln ∣ y ∣ - \frac{3}{4 y} - \frac{1}{16} ln ∣ y + 4 ∣

Add a constant to the solution

= \frac{1}{16} ln ∣ y ∣ - \frac{3}{4 y} - \frac{1}{16} ln ∣ y + 4 ∣ + C

\int (x^{\frac{3}{2}} + 12 x + 2) d x

\int \frac{1}{1 - x + 3 1 - x} d x

\int \frac{6}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

\int \frac{6 x ^{3}}{x ^{4} + 8 x ^{2} + 16} d x

\int \frac{8}{x ln ( x )} d x

What is the integral of ((y+3))/(y^3+4y^2) ?
The integral of ((y+3))/(y^3+4y^2) is 1/16 ln|y|-3/(4y)-1/16 ln|y+4|+C