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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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