What is the integral of (3y^4+9y^2-1)/(y^3+3y) ?

The integral of (3y^4+9y^2-1)/(y^3+3y) is (3y^2)/2-1/3 ln|y|+1/6 ln|y^2+3|+C

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integral of (3y^4+9y^2-1)/(y^3+3y)

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

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

Solution steps

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

Long division

\frac{3 y ^{4} + 9 y ^{2} - 1}{y ^{3} + 3 y} : 3 y - \frac{1}{y ^{3} + 3 y}

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

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

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

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

Simplify

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

Add a constant to the solution

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

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What is the integral of (3y^4+9y^2-1)/(y^3+3y) ?
The integral of (3y^4+9y^2-1)/(y^3+3y) is (3y^2)/2-1/3 ln|y|+1/6 ln|y^2+3|+C