What is the integral of (2y^4+6y^2-1)/(y^3+3y) ?

The integral of (2y^4+6y^2-1)/(y^3+3y) is y^2-1/3 ln|y|+1/6 ln|y^2+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2y^4+6y^2-1)/(y^3+3y)

\int \frac{2 y ^{4} + 6 y ^{2} - 1}{y ^{3} + 3 y} d y

y^{2} - \frac{1}{3} ln ∣ y ∣ + \frac{1}{6} ln y^{2} + 3 + C

Solution steps

\int \frac{2 y ^{4} + 6 y ^{2} - 1}{y ^{3} + 3 y} d y

Long division

\frac{2 y ^{4} + 6 y ^{2} - 1}{y ^{3} + 3 y} : 2 y - \frac{1}{y ^{3} + 3 y}

= \int 2 y - \frac{1}{y ^{3} + 3 y} 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 2 y d y - \int \frac{1}{y ^{3} + 3 y} d y

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

\int \frac{1}{y ^{3} + 3 y} d y = \frac{1}{3} ln ∣ y ∣ - \frac{1}{6} ln y^{2} + 3

= y^{2} - (\frac{1}{3} ln ∣ y ∣ - \frac{1}{6} ln y^{2} + 3)

Simplify

= y^{2} - \frac{1}{3} ln ∣ y ∣ + \frac{1}{6} ln y^{2} + 3

Add a constant to the solution

= y^{2} - \frac{1}{3} ln ∣ y ∣ + \frac{1}{6} ln y^{2} + 3 + C

\int \frac{y ^{3}}{1 + 9 y ^{4}} d y

\int ln (a + x) d x

\int 6 tan (x) sec^{5} (x) d x

\int \frac{y ^{\frac{1}{3}}}{( y ^{\frac{4}{3}} + 9 ) ^{2}} d y

\int \frac{4 x ^{5}}{4 + x ^{2}} d x

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