What is the integral of (y^2+16)/y ?

The integral of (y^2+16)/y is (y^2)/2+16ln|y|+C

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integral of (y^2+16)/y

\int \frac{y ^{2} + 16}{y} d y

\frac{y ^{2}}{2} + 16 ln ∣ y ∣ + C

Solution steps

\int \frac{y ^{2} + 16}{y} d y

Expand

\frac{y ^{2} + 16}{y} : y + \frac{16}{y}

= \int y + \frac{16}{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{16}{y} d y

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

\int \frac{16}{y} d y = 16 ln ∣ y ∣

= \frac{y ^{2}}{2} + 16 ln ∣ y ∣

Add a constant to the solution

= \frac{y ^{2}}{2} + 16 ln ∣ y ∣ + C

\frac{d}{d x} (ln (\frac{x ^{3} - 2 x}{4 + x}))

f (x) = \frac{2 x + 1}{x + 2}

\frac{d}{d x} (\frac{x}{2})

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 9

y = (\frac{2}{( 2 x - 6 ) \frac{3}{2}})

What is the integral of (y^2+16)/y ?
The integral of (y^2+16)/y is (y^2)/2+16ln|y|+C