What is the integral of (3y^4+12y^2-1)/(y^3+4y) ?

The integral of (3y^4+12y^2-1)/(y^3+4y) is 3/2 y^2-1/4 ln|y|+1/8 ln|y^2+4|+6+C

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integral of (3y^4+12y^2-1)/(y^3+4y)

\int \frac{3 y ^{4} + 12 y ^{2} - 1}{y ^{3} + 4 y} d y

\frac{3}{2} y^{2} - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln y^{2} + 4 + 6 + C

Solution steps

\int \frac{3 y ^{4} + 12 y ^{2} - 1}{y ^{3} + 4 y} d y

Expand

\frac{3 y ^{4} + 12 y ^{2} - 1}{y ^{3} + 4 y} : \frac{3 y ^{4}}{y ^{3} + 4 y} + \frac{12 y ^{2}}{y ^{3} + 4 y} - \frac{1}{y ^{3} + 4 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{3 y ^{4}}{y ^{3} + 4 y} d y + \int \frac{12 y ^{2}}{y ^{3} + 4 y} d y - \int \frac{1}{y ^{3} + 4 y} d y

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

\int \frac{12 y ^{2}}{y ^{3} + 4 y} d y = 6 ln y^{2} + 4

\int \frac{1}{y ^{3} + 4 y} d y = \frac{1}{4} ln ∣ y ∣ - \frac{1}{8} ln y^{2} + 4

= \frac{3}{2} (y^{2} + 4 - 4 ln y^{2} + 4) + 6 ln y^{2} + 4 - (\frac{1}{4} ln ∣ y ∣ - \frac{1}{8} ln y^{2} + 4)

Simplify

\frac{3}{2} (y^{2} + 4 - 4 ln y^{2} + 4) + 6 ln y^{2} + 4 - (\frac{1}{4} ln ∣ y ∣ - \frac{1}{8} ln y^{2} + 4) : \frac{3}{2} y^{2} - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln y^{2} + 4 + 6

= \frac{3}{2} y^{2} - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln y^{2} + 4 + 6

Add a constant to the solution

= \frac{3}{2} y^{2} - \frac{1}{4} ln ∣ y ∣ + \frac{1}{8} ln y^{2} + 4 + 6 + C

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What is the integral of (3y^4+12y^2-1)/(y^3+4y) ?
The integral of (3y^4+12y^2-1)/(y^3+4y) is 3/2 y^2-1/4 ln|y|+1/8 ln|y^2+4|+6+C