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

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

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

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

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

Solution steps

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

Long division

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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